Consider the matrix $$ A = \begin{pmatrix} 0 & y & -x\\ y & y^2 & -xy\\ -x & -xy & x^2 \end{pmatrix}. $$ Is it possible to find matrices $X = X(x)$ and $Y=Y(y)$ such that $A = XY$ (or $A = YX$)?
A possibly unrelated observation of mine is that if we consider the vector $v = \begin{pmatrix}y\\-x\end{pmatrix}$, then we can write $A$ in block form as $$ A = \begin{pmatrix} 0 & v^t\\ v & vv^t \end{pmatrix}, $$ which allows us to write $$ A = \begin{pmatrix} 1 & 1\\ 0 & v \end{pmatrix}\cdot \begin{pmatrix} -1 & 0\\ 1 & v^t \end{pmatrix}, $$ but this is not really what I want since now the factors depend on both $x$ and $y$.
EDIT: As suggested in the comments, setting $z = -x$ yields $$ A = \begin{pmatrix} 0 & y & z\\ y & y^2 & yz\\ z & yz & z^2 \end{pmatrix}. $$
Becasue of the symmetry, if it is possible with $A=YX$ then it is possible with $A=XY$ too. So without loss of generality let us assume that we can write $A(x,y)=X(x)Y(y)$ for some matrix-valued functions $X$ and $Y$.
Now setting $x=-1$ we get $$ X(-1)Y(y)\begin{pmatrix}1-w\\0\\w\end{pmatrix} = \begin{pmatrix} 0 & y & 1 \\ y & y^2 & y \\ 1 & y & 1 \end{pmatrix} \begin{pmatrix}1-w\\0\\w\end{pmatrix} = \begin{pmatrix} w \\ y \\ 1\end{pmatrix}$$ and these vectors clearly span $\mathbb R^3$ when we allow $y$ and $w$ to vary. So $X(-1)$ must have full rank.
A similar argument from the other side shows that $Y(1)$ must have full rank too.
But $A(-1,1)$ has rank $2$ and cannot be the product of invertible $X(-1)$ and $Y(1)$.