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 $g(x)h(y)$ ?

 A: For $f(x,y)=\dfrac{xy}{1-x^2y^2}$, the answer is also no.  Suppose on the contrary that there exist nonempty open intervals $U$ and $V$ such that $f|_{U\times V}$ can be factorized as $$f(x,y)=g(x)\,h(y)\text{ for all }x\in U\text{ and }y\in V\,.$$
Fix $a\in U$ and $b\in V$.  Without loss of generality, we may assume that $a$ and $b$ are both nonzero.  Then,
$$\frac{bx}{1-b^2x^2}=f(x,b)=g(x)\,h(b)$$
for all $x\in U$.  This shows that $h(b)\neq 0$ and so there exists $\alpha \neq 0$ such that
$$g(x)=\frac{\alpha x}{1-b^2x^2}\text{ for }x\in U\,.$$
Similarly,
$$h(y)=\frac{\beta y}{1-a^2y^2}\text{ for }y\in V\,.$$
That is,
$$\frac{xy}{1-x^2y^2}=f(x,y)=g(x)\,h(y)=\frac{\alpha \beta\, xy}{\left(1-b^2x^2\right)\,\left(1-a^2y^2\right)}$$
for all $x\in U$ and $y\in V$.  In other words,
$$\left(1-b^2x^2\right)\,\left(1-a^2y^2\right)=\alpha\beta\,\left(1-x^2y^2\right)$$
for all $x\in U\setminus\{0\}$ and $y\in V\setminus\{0\}$.  The two bivariate polynomials $\left(1-b^2x^2\right)\,\left(1-a^2y^2\right)$ and $\alpha\beta\,\left(1-x^2y^2\right)$ must equal (not just as functions, but as polynomials in $\mathbb{R}[x,y]$), but the one on the left has a term $x^2$ with nonzero coefficient, while the one on the right does not have such a term.
A: 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.
A: $$f(x,0)=1=h(0)\cdot h(y)$$ 
and
$$f(0,y)=1=g(x)\cdot h(0)$$
so that both $g(x)$ and $h(y)$ are constant !?
A: 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&=0\\
cb&=0\\
ac&=-1\\
bd&=1
\end{align}
By $bd=1$ we obtain $a=c=0$. Hence this problem has no solution.
