TITLE: Why is this function not differentiable?
QUESTION [5 upvotes]: 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?

REPLY [3 votes]: 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.