Why is this function not differentiable? Why is the function
$$\begin{align}
f(x,y)=\begin{cases}\frac{y^3}{x^2+y^2},&(x,y)\ne(0,0)\\\\
0,&(x,y)=(0,0)
\end{cases}
\end{align}$$
not differentiable at $(0,0)$?
By setting $x=0$ or $y=0$ I would get  $f_x(0,0)=0$, $f_y(0,0)=-2$ and when I look at the 3-D Surface plot, $f(x,y)$ looks quite "normal". I gaze already hours on it and cannot identify the problem.
Obviously the partial derivates exist, but are not continuous, but I cannot see where.
Is there an analytic way, I can obtain the result from? How can I come to the conclusion, without gazing at the curve?
 A: Instead of using theorems about the relationship between differentiability and properties of partial derivatives, let's just go back to the definition of differentiability:
A function $f(x,y)$ is differentiable at $(a,b)$ if and only if there is a linear function $L(x,y)=cx+dy+e$ such that
$$\lim_{(x,y)\to(a,b)}{f(x,y)-L(x,y)\over||(x,y)-(a,b)||}=0$$
For the function $f(x,y)=y^3/(x^2+y^2)$ and $(a,b)=(0,0)$, polar coordinates $x=r\cos\theta$ and $y=r\sin\theta$ gives
$${f(x,y)-L(x,y)\over||(x,y)-(a,b)||}={r\sin^3\theta-cr\cos\theta-dr\sin\theta-e\over r}=\sin^3\theta-c\cos\theta-d\sin\theta-{e\over r}$$
We can get a limit as $r\to0$ by setting $e=0$, but there is no trigonometric identity of the form $\sin^3\theta=c\cos\theta+d\sin\theta$, so the limit as $(x,y)\to(0,0)$ is not only not $0$, it doesn't exist.
Remark: It's not strictly necessary to do so here, but it might help to keep in mind that the choice of $\theta$ can itself vary with $r$, writing it as $\theta_r$ or $\theta(r)$.  Doing so helps emphasize that the limit as $(x,y)\to(0,0)$ is indeterminate.
A: A few comments too long to be put in the comment section.
1: Quote from here: However, the existence of the partial derivatives (or even of all the directional derivatives) does not in general guarantee that a function is differentiable at a point.
2: In your case $f_y(0,0) = 1$, not $-2$.
3: Just because the partial derivatives $f_x$ and $f_y$ are not continuous at $(0,0)$, does not mean $f$ is not differentiable at $(0,0)$. You can use the definition of differentiability to prove it. You can follow this example.
A: You can show that the partial derivatives are not continuous:
$$
\nabla f =
\begin{pmatrix}
-\frac{2 x y^{3} }{(x^{2} + y^{2})^2} \\
\frac{3 x^2 y^2 -y^4}{(x^2+y^2)^2}
\end{pmatrix}
$$
We convert this into polar coordinates:
$$
\nabla f =
\begin{pmatrix}
-2 \cos{\alpha} \sin{\alpha}^3 \\
\sin{\alpha}^2 (\cos{2 \alpha}+2)
\end{pmatrix}
$$
You see that both components depend on the angle, so the partial derivatives are not steady and the function is not differentiable in $(0, 0)^\top$.
