Differentiability of a 2-variable function In this morning's Mathematical Analysis 2 exam, students were asked to study the continuity and differentiability of:
$$f(x,y)=\left\{\begin{array}{cc} \dfrac{xye^{-\frac{1}{(x+y)^2}}}{x^2+2e^{-\frac{2}{(x+y)^2}}} & x\neq-y \\ {} \\ 0 & x=-y \end{array}\right.$$
This question answered the continuity problem. However, I feel if I try the same things I tried there for the differentiability, I will get stuck on the limit involved. Now, it is easy to verify that the partial derivatives in the origin are both zero, since the function is always 0 on the axes. So to prove differentiability, I would have to prove:
$$\lim_{(h,k)\to(0,0)}\frac{f(h,k)}{\sqrt{h^2+k^2}}=0,$$
which is the limit proved zero for continuity except for the denominator, which makes it impossible to use the trick the answer to the linked question used. Polar coordinates give all the same problem as continuity. I can try substituting $k=m|h|^\alpha$, so for $\alpha<\frac12$ I can use the asymptotic in the comments to the linked question, and I'm left with $\frac{m|h|^{\alpha-2}\operatorname{sgn}h}{\sqrt{1+m^2|h|^{2\alpha-2}}}e^{-\frac{1}{m^2|h|^{2\alpha}}}$, and I'm not all too sure that tends to 0. And anyway I would still find problems with $\alpha>\frac12$, and L'Hospital, in this case, is just terrible. Any suggestions?
PS The present question and this one are NOT duplicates, because that one focuses on continuity and this one focuses on differentiability. I thought I had made that clear in this question, but evidently it is not that clear since this question has been marked as a possible duplicate of the continuity one.
 A: First, we prove that the function is differentiable at $p = (a,- a)$ for $a \ne 0$. The function $x^2$ is bounded below in a neighbourhood of $p$, and $xy$ is bounded above. Therefore, letting $x = a + h, y = -a + k$, in order to prove differentiability at $p$ (with derivative $0$), it's enough to prove that
$$\frac{e^{-1/(h + k)^2}}{\sqrt{h^2 + k^2}} \to 0 \text{ as } h,k \to 0.$$
Letting $s = h^2 + k^2$, we have $(h + k)^2 \leq 2s$, so the function under consideration is bounded by $e^{-1/2s}/\sqrt{s}.$ If we set $t = 1/2s$, we see that it's enough to prove $\sqrt{t}e^{-t} \to 0$ as $t \to +\infty$, which is well-known.
Now we prove that $f$ is not differentiable at $(0,0)$. As you mentioned, the partial derivatives at $(0,0)$ are both zero, so if $f$ is differentiable, its derivative must be zero. Therefore the problem is to determine whether 
$$\lim_{x,y \to 0}\frac{1}{\sqrt{x^2 + y^2}} \frac{xye^{-1/(x+y)^2}}{x^2 + 2e^{-2/(x+y)^2}} = 0.$$
However, if we look at the behaviour of this function along the path $(x,y) = (e^{-1/t^2},t-e^{-1/t^2})$ as $t \to 0^{+}$, we find that the limit is $1/3$. The expression within the limit becomes 
$$\frac{1}{3}\left(1 - \frac{e^{-1/t^2}}{t}\right)\frac{1}{\sqrt{1 - 2e^{-1/t^2}/t + 2(e^{-1/t^2}/t)^2}},$$
so to establish that the limit is $1/3$, we just need to see that $e^{-1/t^2}/t$ tends to zero, and this can be established by the same sort of argument as in the previous case.
