Discontinuity of $f(x,y)=\frac{2(x^3+y^3)}{x^2+2y}$ at $(x,y) = (0,0)$ Let $f(x,y)=\frac{2(x^3+y^3)}{x^2+2y}$ , when $(x,y) \neq (0,0)$ and $f(x,y)=0 $ when $(x,y)=(0,0)$ 
We are required to prove the discontinuity of this function at $(0,0)$.
So, I put $y=mx$ where $x \rightarrow 0$, to get the limit to be $0$.
But I can't find another instance where the limiti is different.
Can you guys help?
Thanks

Credit: Carmeister's solution.
I am just curious about the behavior of other similar path.
Let $y = -\frac{x^2}{2}+x^n$ where $n >0$,then we have
\begin{align}\lim_{x \to 0}\frac{2(x^3+(-\frac{x^2}{2}+x^n)^3)}{2x^n} &= \lim_{x \to 0} \frac{(x^3+(-\frac{x^2}{2}+x^n)^3)}{x^n} \\
&= \lim_{x \to 0} \left[ x^{3-n}+\left(-\frac{x^{2+\frac{n}{3}}}2+x^{\frac{4n}{3}} \right)^3\right]\end{align}
If $n \in (0,3)$, the limit is $0$.
If $n>3$, it is undefined.
If $n=3$, \begin{align} \lim_{x \to 0} \left[ x^{3-n}+\left(-\frac{x^{2+\frac{n}{3}}}2+x^{\frac{4n}{3}} \right)^3\right]=1\end{align}
