Rewrite $f(x,y) = 1-x^2y^2$ as a product $g(x) \cdot h(y)$ Rewrite $f(x,y) = 1-x^2y^2$ as a product $g(x) \cdot h(y)$ (both arbitrary functions)

To make more clear what I'm talking about I will give a example.

Rewrite $f(x,y) = 1+x-y-xy$ as $g(x)h(y)$

If we choose $g(x) = (1+x)$ and $h(y) = (1-y)$ we have

$f(x,y) = g(x) h(y) \implies (1+x-y-xy) = (1+x)(1-y)$

I'm trying to do the same with $f(x,y) = 1-x^2y^2 = (1-xy)(1+xy)$.

New question:


  Is there also a contradiction for $f(x,y) = \frac{xy}{1-x^2y^2}$ ? Or it's possible to write $f(x,y) $ as $g(x)h(y)$ ?
Suppose we had functions $g,h$ with $g(x) h(y) = 1 - x^2 y^2$.  Then substituting in $x:=1, y:=1$, we would have $g(1) h(1) = 0$, so either $g(1) = 0$ or $h(1) = 0$.  But we also must have $g(1) h(0) = 1$ which implies $g(1) \ne 0$, and similarly $g(0) h(1) = 1$ which implies $h(1) \ne 0$.  Putting these requirements together, we get a contradiction.

Therefore, no such functions $g,h$ can exist.
Without loss of generality let $f(x) = ax^2 + b$ and $g(y) = cy^2 + d$. Now
$$
1-x^2y^2 = f(x)g(y) = acx^2y^2 +adx^2 = cby^2 +bd
$$
By comparing terms we obtain:
\begin{align}
ad&amp;=0\\
cb&amp;=0\\
ac&amp;=-1\\
bd&amp;=1
\end{align}
By $bd=1$ we obtain $a=c=0$. Hence this problem has no solution.