At what points would $f$ be differentiable?

$f(x_1,x_2)=\begin{cases} \dfrac{x_1^3}{|x_1|+|x_2|} \text{ for } (x_1,x_2)\neq (0,0)\\ 0\text { if }(x_1,x_2)=(0,0) \end{cases}$

Only place that I can think of where any trouble could possibly arise is in the denominator, where we have the sum of two absolute value functions. From the single variable case, we know that $f(x)=|x|$ lacks differentiability at the origin, but in this function, we don't have to consider the origin since we are given that $(x_1,x_2)\neq (0,0)$.

Any help would be appreciated. Wondering if I'm overthinking this...


Some hints: $f$ is $C^\infty$ in each of the open quadrants, so $f$ is certainly differentiable there. That leaves the axes. Along the $y$ axis we can use this: If $f(a,b) = 0$ and $f(x,y) = o(|(x,y)-(a,b)|)$ as $(x,y)\to (a,b),$ then $f$ is differentiable at $(a,b).$ (In fact $Df(a,b)$ is just the $0$ linear transformation.) On the $x$ axis, $x\ne 0:$ Here you can expect some trouble. Take $(1,0)$ for example, and consider $f(1,y) = 1/(1+|y|).$ That $|y|$ doesn't bode well for $\partial f /\partial y$ at $(1,0),$ never mind the total derivative there.


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