Show limit exists if and only if $m+n>2$. Let $m,n \in \mathbb{N}$.
Show that the limit $\lim_{x \to (0,0)} \frac{x^ny^m}{x^2+y^2}$ exists if an only if $n+m>2$.
Attempt:
$(\Leftarrow)$. If $n+m>2$ without loss of generality assume $n \geq m$. Then $|\frac{x^ny^m}{x^2+y^2}|\leq |\frac{x^ny^m}{x^2}|=|x^{n-2}y^m|\leq (\sqrt{x^2+y^2})^{n-2}(\sqrt{x^2+y^2})^m=(\sqrt{x^2+y^2})^{m+n-2}$.
So for $\epsilon>0$, choose $\delta=\epsilon^{\frac{1}{m+n-2}}$, then $0<\sqrt{x^2+y^2}<\delta \implies |\frac{x^ny^m}{x^2+y^2}|<\epsilon$.
$(\Rightarrow)$.If $n+m\leq 2$. $f(\frac{1}{k},\frac{1}{k})=(\frac{1}{2})(\frac{1}{k})^{n+m-2}$ which does not approach zero as $k \rightarrow 0$, while $f(\frac{1}{k},0)=0$ approaches $0$ as $k \rightarrow 0$. Since these limits are different, the limit does not exist in this case.

Your proof seems to be correct! Another more intuitive way would be to use polar coordinates in this situation (which always is a good thing to consider when you have a $x^2+y^2$ term somewhere): $$\lim \limits_{(x,y) \to 0} \frac{x^ny^m}{x^2+y^2}=\lim \limits_{r\to 0}\frac{(r\cos(\theta))^n(r\sin(\theta))^m}{(r\cos(\theta))^2+(r\sin(\theta))^2}=\lim \limits_{r\to 0} r^{n+m-2}\cos(\theta)^n\sin(\theta)^m \rightarrow 0 \quad \text{when} \; n+m>2$$
(since $cos(\theta)^n\sin(\theta)^m$ is bounded)
