Let $V$ be a vector space over $\Bbb F$, and let $x\not=0,y\not=0 $ be two elements in $V$.
I want to show that $x\otimes_{_F} y=y\otimes_{_F} x$ iff $x=ay$ where $a\in \Bbb F$.
I know the second direction, so only want to see the first direction (If case).
So for such $x,y\neq 0$, you want to show $x\otimes y=y\otimes x$ in $V\otimes_F V$ if $x=ay$ for some $a\in F$.
Then $x\otimes y=ay\otimes y=y\otimes ay=y\otimes x$.
Since this seems by far to be the easier half of the problem, I am beginning to wonder if you meant to ask about the other direction.
Suppose $x,y$ are linearly independent. As such, this pair can be extended with other elements of $V$ to form a basis $\beta$ of $V$. We know that given a basis $\{b_i\mid i\in I\}$ for $V$, we automatically have a basis $\{b_i\otimes b_j\mid i,j\in I\}$ for $V\otimes_F V$.
Applying this to our basis $\beta$, we have that $x\otimes y$ and $y\otimes x$ are linearly independent elements of a basis of $V\otimes_F V$, and so certainly $x\otimes y\neq y\otimes x$.
By proving the contrapositive, we've shown that if $x\otimes y=y\otimes x$, then $x,y$ are linearly dependent, hence $x=ay$ for some $a\in F$.
Take a set of basis $e_1,\ldots,e_n$ of $V$ and let $x=\sum_ {i}x^ie_i,~y=\sum_ {j}y^je_j$, then $$x\otimes y=\sum_ {i,j}(x^ie_i)\otimes(y^je_j)=\sum_ {i,j}x^iy^je_i\otimes e_j$$ $$y\otimes x=\sum_ {i,j}y^jx^ie_j\otimes e_i$$ The symmetry implies $$x^iy^j=x^jy^i$$ That is, $$\frac{x^i}{y^i}=\frac{x^j}{y^j}=a$$ for some constant $a$.
