TITLE: Differentiability of a function $\mathbb{R}^2\to\mathbb{R}$ at $(0,0)$
QUESTION [5 upvotes]: I am trying to solve this question from a past paper.
Determine how the differentiability of $f(x,y)$ below depends on the real parameter $\alpha > 0$
$$f(x,y)=\left\{ \begin{array}{ll} \displaystyle \frac{\log(1+|x|^\alpha y^2)}{x^2+y^4} & \text{if } (x,y) \neq (0,0) \\ 0 & \text{if } (x,y)  = (0,0)\end{array} \right.$$
It's easy to see that $f$ is differentiable at all $(x,y) \neq (0,0) $ for all $\alpha >0$.
So we need to study what happens at $(0,0)$.
As we have $\frac{\partial f}{\partial x}(0,0)=\frac{\partial f}{\partial y}(0,0)=0$, the definition of differentiability reduces to showing that the limit:
$$\lim_{(x,y)\to(0,0)} \frac{\log(1+|x|^\alpha y^2)}{\sqrt{x^2+y^2}(x^2+y^4)}  $$
equals $0$.
Using Young's inequality for products, we know this is the case for all $\alpha > 2$.
On the curve $x=y^2$, the above limit is not $0$ for $\alpha < \tfrac{3}{2}$.
I am not sure how to proceed from here. I have tried other paths, but to no avail.
I would appreciate any tips or pointers.

REPLY [3 votes]: The function will be differentiable whenever $\alpha>\frac32$. Here are a few hints: Use the first-order Taylor approximation of $\log(1+u)$ (this I assume you already did), and then think about a modified polar coordinates substitution $x=r\cos\theta$, $y^2=r\sin\theta$.
It's sort of fun, more generally, to consider the function
$$f(x,y)=\begin{cases} \frac{|x|^\alpha|y|^\beta}{|x|^\gamma+|y|^\delta}, & (x,y)\ne (0,0) \\ 0\, & (x,y)=(0,0)\end{cases}$$
and determine for what values of $\alpha,\beta,\gamma,\delta$ it is continuous and then differentiable.