Where is this function from $\mathbb{R}^2 \to \mathbb{R}$ differentiable? $$
f(x,y) = \frac{x^3+2y^4}{x^2+y^2}.
$$
and zero for $ x=y=0.$
What is the general approach for problems like these? Clearly the function is differentiable at every point that is not the origin, but here it is not clear to me what to do.
I can find the partial derivatives and this is just a rational function in $x,y.$ Not sure what this helps me with though.
Thanks a lot in advance!
 A: You can use here polar coordinates:
$x=r\cos\theta$, $y=r\sin\theta$.
Then our given function becomes
$f(r,\theta)=\frac{r^3\cos\theta+2r^4\sin\theta}{r^2}$
$f(r,\theta)=r\cos\theta+2r^2\sin\theta$
Now, when $x,y\to 0$, $r\to 0$
$\displaystyle\lim_{r\to 0}\frac{f(r,\theta)-f(0,\theta)}{r}
=\displaystyle\lim_{r\to 0}(\cos\theta+2r\sin\theta)$.
Now, limit is not unique, it depends on $\theta$.
Basically limit depends upon choice of path we follow to approach the origin $(0,0)$.
Hence $f$ is not differentiable at $(0,0)$.
A: To be differentiable at the origin, there must be a constant vector $\mathbf{J}$ such that
$$
\lim_{|\mathbf{h}|\rightarrow 0}\frac{f(h_x,h_y) - f(0,0) - \mathbf{J}\cdot\mathbf{h}}{|\mathbf{h}|} = 0.
$$
Now, let $\mathbf{h} = (h\cos\theta,h\sin\theta)$. We have
\begin{eqnarray}
\lim_{|\mathbf{h}|\rightarrow 0}\frac{f(h_x,h_y) - f(0,0) - \mathbf{J}\cdot\mathbf{h}}{|\mathbf{h}|} &=& \lim_{|\mathbf{h}|\rightarrow 0}\frac{1}{h}\left(\frac{h^3\cos^3\theta+2h^4\sin^4\theta}{h^2\cos^2\theta+h^2\sin^2\theta}-J_xh\cos\theta - J_yh\sin\theta\right) \\&=& \lim_{|\mathbf{h}|\rightarrow 0}\left[\cos^3\theta +2h\sin^4\theta - J_x\cos\theta - J_y\sin\theta\right]
\end{eqnarray}
There are no constants $J_x$, $J_y$ that can make that limit $0$ independent of $\theta$, so the function is not differentiable at the origin.
