Find all non-constant function $F(x, y)\in C^2(\mathbb{R}^2)$ such that under the Cartesian coordinate system $F(x, y) = f(x) g(y)$ but under the polar coordinate system $F(x, y) = h(r)$.
My thoughts: setting x=0 and y=0 we get $F(x,0)=F(0,x) = h(|x|)$. So $f(0)g(x)=f(x)g(0)$. WLOG, $g(0)=1$. Now taking partial derivative w.r.t $\theta$ in $F(r\cos\theta,r\sin\theta)=h(r)$, we get
$\frac{g'(x)}{xg(x)}=\frac{f'(x)}{xf(x)}=\frac{g'(y)}{yg(y)}$ where $x=r\cos\theta,y=r\sin\theta$. So this equality is true for any $(x,y)$ with $x^2+y^2=r^2$.
But if we suppose $\frac{g'(x)}{xg(x)}=\frac{g'(y)}{yg(y)}$ is true for any $x,y\in[-r,r]$, then we can set it equal to some constant $c_0$ and solve the differential equation. And we should get $F(x,y)=k\exp(c(x^2+y^2))$. But the challenge is to show this form of solution of unique.
It looks as if you have already solved the problem, but didn't realise this.
Your argument shows that there exists an absolute (i.e. dependent only on $f,g,h$) constant $c_0$ such that $f'(x)/x f(x) = c_0$. (You don't really have to worry about the constraint $x,y \in [-r,r]$ - just argue that changing $r$ does not change $c_0$, and then take any $r$ large enough.)
It follows that $f(x) = k_1 e^{c x^2}$ with $c = c_0/2$, and $g(x) = k_2 e^{c y^2}$, and finally $F(x,y) = k e^{c x^2 + y^2} = k e^{r^2}$.
The above argument shows that: if $F$ satisfies the assumptions, then it is of the form $F(x,y) = k e^{c x^2 + y^2}$ - no other $F$ are under consideration. Now, of course, any function of the above form satisfies the assumptions - so, you have indeed classified them all.