Q:

Showing $f$ is not differentiable at $(0,0)$ even though the partial derivatives exist.

$f\begin{pmatrix}x\\y\end{pmatrix}:=\begin{cases}\frac{x^2y}{x^6+2y^2}& (x,y)^T\neq(0,0)^T\\0&(x,y)^T=(0,0)^T\end{cases}$

I want to find out if $f$ is continuously partially differentiable at $(0,0)^T$. 

The partial derivates at $(0,0)^T$ are both equal to $0$. So if $f$ were differentiable at $(0,0)^T$ we would have 
$$\lim_{x,y\to 0}\frac{f(x,y)}{\sqrt{x^2+y^2}}=0$$
Using polar coordinates I end up with 
$$\lim_{r\to 0}\frac{r^3 \cos^2 \phi \sin \phi}{(r^6 \cos^6 \phi +2 r^2 \sin^2\phi)r}$$
Which does't converge to $0$. So this means $f$ is not differentiable at $(0,0)^T$ and therefore not continuously partially differentiable at $(0,0)^T$.
Is that correct?

A:

When you wrote$$\lim_{x,y\to 0}\frac{f(x,y)}{\sqrt{x^2+y^2}},$$you meant$$\lim_{x,y\to 0}\frac{f(x,y)}{\sqrt{x^2+y^2}}=0.$$Otherwise, it is correct.

