How to prove this function isn't differentiable at origin [Exercise 5.18 - Pugh] Show that the function $f:\mathbb{R}^2\rightarrow \mathbb{R}$ defined by
$$ f(x,y)= \begin{cases} \frac{x^3y}{x^4+y^2} \ ; (x,y)\neq 0 \\\quad 0\quad\ ;(x,y)=0 \end{cases}$$
has $\nabla_{(0,0)}f(u)=0$ for all $u$ but is not differentiable at $(0,0)$.
My attempt Isn't hard to show the first part (that all directional derivatives exist) but I cannot show that this function is differentiable. My ideia is standard, I need to find a path such that $\lim\limits_{\gamma\;\mapsto (0,0)} f(\gamma)$ doens't exist. In order to achieve this I tried $y=mx, y=mx^2, y=mx^3$ but no one gave me a clear result. At best, I was able to show that
$$\lim\limits_{(x,y)\mapsto(0,0)}\frac{f(x,y=x^2)}{f(x,y=-x^2)}=-1 $$
But I don't think that's enough. Is that enough ? If yes, why? If not, is there a better method to solve this ?
 A: We need to check if $$\frac{x^2y}{x^4+y^2} \overset?= 0 + 0x + 0y + o\left(\sqrt{x^2+y^2}\right) \\
\lim_{(x,y)\to(0,0)} \frac{x^2y}{(x^4+y^2)\sqrt{x^2+y^2}} \overset ? = 0$$
but when $y=kx$
$$\lim_{x\to0}\frac{kx^3}{x^2(x^2+k^2)x\sqrt {1+k^2}} = \frac{1}{k\sqrt{1+k^2}}\ne 0$$
A: Let  $\gamma :\mathbb{R}^{2} \rightarrow  \mathbb{R}^{2},~ \left( x,y \right) \mapsto \left( r\cos  \vartheta ,~r\sin  \vartheta  \right) .$  Then, it follows that
$$
f\circ \gamma ≔f \left(  \gamma  \left( x,y \right)  \right) =f \left( r\cos  \vartheta ,r\sin  \vartheta  \right) =\frac{r^{4}\cos ^{3} \vartheta \sin  \vartheta }{r^{4}\cos ^{4} \vartheta +r^{2}\sin ^{2} \vartheta }=\frac{r^{4}\cos ^{3} \vartheta \sin  \vartheta }{r^{4} \left( \cos ^{4} \vartheta +\frac{1}{r^{2}}\sin ^{2} \vartheta  \right) }=\frac{\cos ^{3} \vartheta \sin  \vartheta }{\cos ^{4} \vartheta +\frac{1}{r^{2}}\sin ^{2} \vartheta } \
$$
Then, if we compute the limit, it is easy to see that it does converge to $0$. So why this method doesn't work? I'm a bit lost.
