GolfScript, 19 characters

n%{'aeiouy'&,6=},n*

Usage:

golfscript vowels.gs < wordlist.txt

Output:

abstemiously
authoritatively
behaviourally
[...]
uninformatively
unintentionally
unquestionably
unrecognisably

If you also want to output the count at the end you can use

n%{'aeiouy'&,6=},.n*n@,

which is four characters longer.

APL, 21 - 5 = 16

(,≢)w/⍨∧⌿'aeiouy'∘.∊w

Expects to find the list of words as w. Returns a list of the words that contain all the vowels, plus their count. Tested with ngn apl. Here's an example.

Explanation

         'aeiouy'∘.∊w   # generate a truth table for every vowel ∊ every word
       ∧⌿               # reduce with ∧ along the rows (vowels)
    w/⍨                 # select the words that contain all the vowels
(,≢)                    # hook: append to the list its own tally

Ruby 38

Edit 34: Best so far (from @O-I):

a.split.select{|w|'aeiouy'.count(w)>5} 

Edit 1: I just noticed the question asked for 'y' to be included among the vowels, so I've edited my question accordingly. As @Nik pointed out in a comment to his answer, "aeiouy".chars is one character less than %w[a e i o u y], but I'll leave the latter, for diversity, even though I'm risking nightmares over the opportunity foregone.

Edit 2: Thanks to @O-I for suggesting the improvement:

s.split.select{|w|'aeiouy'.delete(w)==''}

which saves 11 characters from what I had before.

Edit 3 and 3a: @O-I has knock off a few more:

s.split.select{|w|'aeiouy'.tr(w,'')==''}

then

s.split.reject{|w|'aeiouy'.tr(w,'')[1]}

and again (3b):

a.split.select{|w|'aeiouy'.count(w)>5} 

I am a mere scribe!

Here are two more uncompettive solutions:

s.split.group_by{|w|'aeiouy'.delete(w)}['']       (43)
s.gsub(/(.+)\n/){|w|'aeiouy'.delete(w)==''?w:''} (48)

Initially I had:

s.split.select{|w|(w.chars&%w[a e i o u y]).size==6}

s is a string containing the words, separated by newlines. An array of words from s that contains all five vowels is returned. For readers unfamiliar with Ruby, %w[a e i o u y] #=> ["a", "e", "i", "o", "u", "y"] and & is array intersection.

Suppose

s = "abbreviations\nabduction\n"
s.split #=> ["abbreviations", "abduction"] 

In the block {...}, initially

w             #=> "abbreviations"
w.chars       #=> ["a", "b", "b", "r", "e", "v", "i", "a", "t", "i", "o", "n", "s"]
w.chars & %w[a e i o u y] #=> ["a", "e", "i", "o"]
(w.chars & %w[a e i o u y]).size == 6 #=> (4 == 6) => false,

so "abbreviations" is not selected.

If the string s may contain duplicates, s.split.select... can be replaced by s.split.uniq.select... to remove duplicates.

Just noticed I could save 1 char by replacing size==6 with size>5.

Python - 46 characters

filter(lambda x:set(x)>=set("AEIOUY"),open(f))

Readable version: It's already pretty readable :-)

Haskell - 67

main=interact$unlines.filter(and.flip map "aeiouy".flip elem).lines

Ruby - 28 characters (or 27 if y is excluded from vowels)

("ieaouy".chars-s.chars)==[]

The complete command to run is (48 chars):

ruby -nle 'p $_ if ("ieaouy".chars-$_.chars)==[]'

EDIT: replaced puts with p as suggested by @CarySwoveland

AWK - 29

/a/&&/e/&&/i/&&/o/&&/u/&&/y/

To run: Save the lowercase word list to wordlist.txt. Then, do:

mawk "/a/&&/e/&&/i/&&/o/&&/u/&&/y/" wordlist.txt

If your system does not have mawk, awk can be used as well.

You can also run it from a file by saving the program to program.awk and doing mawk or awk -f program.awk.

k [22-5=17 chars]

I have renamed the file "corncob_lowercase.txt" to "w"

Count the words [22 chars]

+/min'"aeiouy"in/:0:`w

Output

43

Find all words [25 chars]

x@&min'"aeiouy"in/:x:0:`w

Overall 43 words containing all the vowels (a e i o u y)

Output

"abstemiously"
"authoritatively"
"behaviourally"
..
..
"unintentionally"
"unquestionably"
"unrecognisably"

Javascript/JScript 147(152-5), 158(163-5) or 184(189-5) bytes:

Here is my Javascript and JScript horribly "ungolfyfied" version (164 152 152-5=147 bytes):

function(s,k,z,x,i,c,r){c='aeiouy'.split('');r=[];for(k in s=(s+'').split(/\b/)){i=0;for(z in c)i+=s[k].indexOf(c[z])>=0;i==6&&(r[r.length]=s[k]);}return r;}

function(s,k,z,x,i,c,r){c='aeiouy'.split('');r=[];for(k in s=(s+'').split(/\b/)){i=6;for(z in c)i-=!!s[k].search(c[z]);i&&(r[r.length]=s[k]);}return r;}

Thank you @GaurangTandon for the search() function, which saved me a byte!

RegExp based with HORRIBLE performance, but support both upper and lowercase (163-5=158 bytes):

function(s,k,z,x,i,c,r){c='aeiouy'.split('');r=[];for(k in s=(s+'').split(/\b/)){i=0;for(z in c)i+=RegExp(c[z],'i').test(s[k]);i==6&&(r[r.length]=s[k]);}return r;}

RegExp based with BETTER performance, BUT takes a lot more bytes (189-5=184 bytes):

function(s,k,z,x,i,c,r,l){l=[];r=[];for(z in c='aeiouy'.split(''))l[z]=RegExp(c[z],'i');for(k in s=(s+'').split(/\b/)){i=0;for(z in c)i+=l[z].test(s[k]);i==6&&(r[r.length]=s[k]);}return r;}

This one if just for the fun (175-5 bytes) and won't count as an answer:

function(s,k,z,x,i,c,r){c='aeiouy'.split('');r=[];for(k in s=(s+'').split(/\b/)){i=0;for(z in c)i+=s[k].indexOf(c[z])>=0;i==6&&(r[r[r.length]=s[k]]=1+(r[s[k]]||0));}return r;}

It's based on the 1st answer, but has a 'twist': You can know how many times a word has been found.

You simply do like this:

var b=(function(s,k,z,x,i,c,r){c='aeiouy'.split('');r=[];for(k in s=(s+'').split(/\b/)){i=0;for(z in c)i+=s[k].indexOf(c[z])>=0;i==6&&(r[r[r.length]=s[k]]=1+(r[s[k]]||0));}return r;})('youaie youaie youaie youaie a word');

b.youaie //should be 4

Since that length doesn't have all vowels, it wont be deleted and still would be an answer for the bonus.

How do you call it?

"Simple": You wrap the function inside () and then add ('string goes here'); to the end.

Like this: (function(s,k,z,x,i,c,r){c='aeiouy'.split('');r=[];for(k in s=(s+'').split(/\b/)){i=0;for(z in c)i+=s[k].indexOf(c[z])>=0;i==6&&(r[r.length]=s[k]);}return r;})('a sentence youyoy iyiuyoui yoiuae oiue oiuea');

This example will return an array only with 1 string: yoiuae

I know that this is the worst solution, but works!

Why am i counting -5?

Well, Javascript/JScript arrays have a property (length) in arrays which tells the number of elements that it have.

After being confirmed in the question, the bonus of -5 is for telling the number of words.

Since the number of words is in that property, automatically I have the score of -5.

Ruby 39 38

Currently the shortest Ruby entry when counting the whole program including input and output.

Saved a char by using map instead of each:

$<.map{|w|w*5=~/a.*e.*i.*o.*u/m&&p(w)}

Another version, 39 characters with prettier output:

puts$<.select{|w|w*5=~/a.*e.*i.*o.*u/m}

Both programs take input from stdin or as a file name passed as a command line argument:
$ ruby wovels.rb wordlist.txt

It costs 3 extra characters to inclyde y as a wovel.

Mathematica (65 or 314)

Two very different approaches, the better one was proposed by Belisarius in the comments to my initial response. First, my brutish effort, which algorithmically generates every possible regular expression that matches a combination of six vowels (including "y"), and then checks every word in the target wordlist against every one of these 720 regular expressions. It works, but it's not very concise and it's slow.

b = Permutations[{"a", "e", "i", "o", "u","y"}]; Table[
 Select[StringSplit[
  URLFetch["http://www.mieliestronk.com/corncob_lowercase.txt"]], 
  StringMatchQ[#, (___ ~~ b[[i]][[1]] ~~ ___ ~~ 
      b[[i]][[2]] ~~ ___ ~~ b[[i]][[3]] ~~ ___ ~~ 
      b[[i]][[4]] ~~ ___ ~~ b[[i]][[5]]~~ ___ ~~ b[[i]][[6]] ~~ ___)] &], {i, 1, 
  Length[b]}]

~320 characters. A few could be saved through using alternate notation, and additional characters are lost preparing the dictionary file as a list of strings (the natural format for the dictionary in Mathematica. Other languages may not need this prep, but Mathematica does). If we omit that step, presuming it to have been handled for us, the same approach can be done in under 250 characters, and if we use Mathematica's built-in dictionary, we get even bigger savings,

Map[DictionaryLookup[(___ ~~ #[[1]] ~~ ___ ~~ #[[2]] ~~ ___ ~~ #[[3]] 
~~ ___ ~~ #[[4]] ~~ ___ ~~ #[[5]]~~ ___ ~~ #[[6]] ~~ ___) .., IgnoreCase -> True] &, 
 Permutations[{"a", "e", "i", "o", "u","y"}]]

Under 200 characters. Counting the number of words found requires only passing the result to Length[Flatten[]], which can be added around either block of code above, or can be done afterwards with, for example, Length@Flatten@%. The wordlist specified for this challenge gives 43 matches, and the Mathematica dictionary gives 64 (and is much quicker). Each dictionary has matching words not in the other. Mathematica finds "unprofessionally," for example, which is not in the shared list, and the shared list finds "eukaryotic," which is not in Mathematica's dictionary.

Belisarius proposed a vastly better solution. Assuming the wordlist has already been prepared and assigned to the variable l, he defines a single test based on Mathematica's StringFreeQ[] function, then applies this test to the word list using the Pick[] function. 65 characters, and it's about 400 times faster than my approach.

f@u_ := And @@ (! StringFreeQ[u, #] & /@ Characters@"aeiouy"); Pick[l,f /@ l]

Perl 6 - 35 characters

Inspired by @CarySwoveland 's Ruby solution:

say grep <a e i o u y>⊆*.comb,lines

This selects (greps) each line that returns True for <a e i o u y> ⊆ *.comb, which is just a fancy way of asking "is the Set ('a','e','i','o','u','y') a subset (⊆) of the Set made up of the letters of the input (*.comb)?"

Actually, both <a e i o u y> and *.comb only create Lists: ⊆ (or (<=) if you're stuck in ASCII) turns them into Sets for you.

To get the number of lines printed, this 42 character - 5 = 37 point script will output that as well:

say +lines.grep(<a e i o u y>⊆*.comb)».say

Javascript - Score = 124 - 5 = 119 !!!

Edit: 17/02/14

function(l){z=[],l=l.split("\n"),t=0;while(r=l[t++]){o=0;for(i in k="aeiouy")o+=!~r.search(k[i]);if(o==0)z.push(r)}return z}

Big thanks to @Ismael Miguel for helping me cut off ~12 chars!

I removed the fat arrow notation function form because though I have seen it begin used, it doesn't work. No idea why...

To make it work:

Pass all the words separated by newline as an argument to the function as shown below.

Test:

// string is "facetiously\nabstemiously\nhello\nauthoritatively\nbye"

var answer = (function(l){z=[],l=l.split("\n"),t=0;while(r=l[t++]){o=0;for(i in k="aeiouy")o+=!~r.search(k[i]);if(o==0)z.push(r)}return z})("facetiously\nabstemiously\nhello\nauthoritatively\nbye")

console.log(answer);
console.log(answer.length);

/* output ->    
    ["facetiously", "abstemiously", "authoritatively"]
    3 // count
*/

Python, 45 characters

[w for w in open(f) if set('aeiouy')<=set(w)]

Bash (grep) - (36-5 bonus) = 31 bytes

g=grep;$g y|$g u|$g o|$g i|$g a|$g e

Note the order of vowels tested, least frequent first. For the test case, this runs about 3 times as fast as testing in order a e i o u y. That way the first test removes a larger number of words so subsequent tests have less work to do. Obviously this has no effect on the length of the code. Many of the other solutions posted here would benefit similarly from doing the tests in this order.

sed 29 chars

/y/{/u/{/o/{/i/{/a/{/e/p}}}}}

Order choosed from Letter frequency on wikipedia to speed check.

On my host:

time sed -ne '/a/{/e/{/i/{/o/{/u/{/y/p}}}}}' </usr/share/dict/american-english >/dev/null 
real    0m0.046s

and

time sed -ne '/y/{/u/{/i/{/o/{/a/{/e/p}}}}}'</usr/share/dict/american-english >/dev/null 
real    0m0.030s

D - 196

import std.regex,std.stream;void main(string[]a){auto f=new File(a[1]);while(!f.eof)if(!(a[0]=f.readLine.idup).match("(?=.*a)(?=.*e)(?=.*i)(?=.*o)(?=.*u)(?=.*y).*").empty)std.stdio.writeln(a[0]);}

Un-golfed:

import std.regex, std.stream;

void main( string[] a )
{
    auto f = new File( a[1] );

    while( !f.eof )
        if( !( a[0] = f.readLine.idup ).match( "(?=.*a)(?=.*e)(?=.*i)(?=.*o)(?=.*u)(?=.*y).*" ).empty )
            std.stdio.writeln( a[0] );
}

Usage: C:\>rdmd vowels.d wordlist.txt

wordlist.txt must contain the lowercase list words.

Rebol (104 chars)

remove-each w d: read/lines %wordlist.txt[6 != length? collect[foreach n"aeiouy"[if find w n[keep n]]]]

Un-golfed:

remove-each w d: read/lines %wordlist.txt [
    6 != length? collect [foreach n "aeiouy" [if find w n [keep n]]]
]

d now contains list of found words. Here is an example from Rebol console:

>> ; paste in golf line.  This (REMOVE-EACH) returns the numbers of words removed from list

>> remove-each w d: read/lines %wordlist.txt[6 != length? collect[foreach n"aeiouy"[if find w n[keep n]]]]
== 58067

>> length? d
== 43

Smalltalk (36/57 chars)

'corncob_lowercase.txt' asFilename contents select:[:w | w includesAll:'aeiouy']

to get the count, send #size to the resulting collection. The result collection contains 43 words ('abstemiously' 'authoritatively' ... 'unquestionably' 'unrecognisably')

The code above has 77 chars, but I could have renamed the wordlist file to 'w', so I count the filename as 1 which gives a score of 57.

Is reading the file part of the problem or not? If not (see other examples), and the list of words is already in a collection c, then the code reduces to:

c select:[:w | w includesAll:'aeiouy']

which is 36 chars (with omittable whitespace removed).

Mathematica - 136 102

Fold[Flatten@StringCases@## &, 
Flatten@Import@"http://bit.ly/1iZE9kY",
___ ~~ # ~~ ___ & /@ Characters@"aeiouy"]

updated: unnecessary spaces removed

Very slow but in bash (81 chars):

while read l;do [ `fold -w1<<<$l|sort -u|tr -dc ieaouy|wc -m` = 6 ]&&echo $l;done

EDIT: echo $l|fold -w1 replaced with fold -w1<<<$l as suggested by @nyuszika7h

C - 96 bytes

 char*gets(),*i,j[42];main(p){while(p=0,i=gets(j)){while(*i)p|=1<<*i++-96;~p&35684898||puts(j);}}

I saved several bytes of parentheses thanks to a fortunate coincidence of operator precedence.

JavaScript - 95 bytes

Here's my golf.

function f(s){return[var a=s.match(/^(?=.*a)(?=.*e)(?=.*i)(?=.*o)(?=.*u)(?=.*y).*/),a.length];}

And I would also like to point out that your golf doesn't seem to find all occurrences of vowels.

Ungolfed:

function countVowels(string) {
  var regex   = /^(?=.*a)(?=.*e)(?=.*i)(?=.*o)(?=.*u)(?=.*y).*/;
  var matches = string.match(regex);
  var length  = matches.length;
  return [matches, length];

Bash + coreutils, 39

eval cat`printf "|grep %s" a e i o u y`

Takes input from stdin.

sort + uniq + sed

This one does not match repeated occurrences of a word. It also does not match the letter 'y' if is occurs at the beginning of a word.

sort wordlist.txt | uniq | sed -n '/a/p' | sed -n '/e/p' | sed -n '/i/p' | sed -n '/o/p' | sed -n '/u/p' | sed -nr '/^[^y]+y/p' 

Bash

Not as short as the OP's , but one line in Bash:

while read p; do if [ $(sed 's/[^aeiouy]//g' <<< $p | fold -w1 | sort | uniq | wc -l) -eq 6 ] ; then echo $p; fi; done < wordlist.txt

C-Sharp

I've never done this before and I'm not exactly sure what the posting procedures are. But this is what i came up with:

Action<string>x=(stream=>stream.Split('\n').ToList().ForEach(word=>{if(new char[]{'a','e','i','o','u','y'}.All(v=>word.ToCharArray().Contains(v)))Console.Write(word);}));using(var stream=new StreamReader("corncob_lowercase.txt"))x(stream.ReadToEnd());

wordList = a List<string> of all the words.

if you want to display a total:

Action<string>x=(stream=>{var total=0;stream.Split('\n').ToList().ForEach(word=>{if(new char[]{'a','e','i','o','u','y'}.All(v=>word.ToCharArray().Contains(v))){Console.Write(word);total+=1;}});Console.Write(total);});using(var stream=new StreamReader("corncob_lowercase.txt"))x(stream.ReadToEnd());

// init
var words = "";
var vowels = new char[] {'a','e','i','o','u','y'};
var total = 0;

using (var stream = new StreamReader("corncob_lowercase.txt"))
{
    // read file
    words = stream.ReadToEnd();

    // convert word to List<string>
    var wordList = words.Split('\n').ToList();

    // loop through each word in the list
    foreach (var word in wordList)

        // check if the current word contains all the vowels
        if(vowels.All (w => word.ToCharArray().Contains(w)))
        {
            // Count total
            total += 1;
            // Output word
            Console.Write(word);
        }

    // Display total
    Console.WriteLine(total);
}

Batch - 52 Bytes

if you're not worried about the original word list -

@for %%a in (a e i o u)do @find "%%a" %1>f&type f>%1

This outputs the new list to a file called f, preserving the original word list - 66 bytes -

@type %1>f&for %%a in (a e i o u)do @find "%%a" f>f%%a&type f%%a>f

If you want it to clean up the files it leaves around, just add &del f%%a to the end for an extra 9 bytes.

Takes list of words as input.

H:\uprof>vowles.bat wordlist.txt

Old solution - 162 Bytes

@echo off&setLocal enableDelayedExpansion&for /f %%a in (%1) do (set a=%%a&set t=&for %%b in (a e i o u) do if "!a:%%b=!"=="!a!" set t=1)&if "!t!" NEQ "1" echo %%a

JavaScript - 118 bytes

function(a){r=[];for(k in a=a.split("\n")){c=0;for(b in s="aeiouy")c+=!!~a[k].search(s[b]);c>5&&r.push(a[k])}return r}

Paste this into the console to test it:

(function(a){r=[];for(k in a=a.split("\n")){c=0;for(b in s="aeiouy")c+=!!~a[k].search(s[b]);c>5&&r.push(a[k])}return r})(document.getElementsByTagName('pre')[0].innerText)