Is $f(x,y) = \frac{xy}{(x^2 + y^2)^{\frac{1}{3}}}$ differentiable at $(0,0)$?

Question

We have $f(x,y) = \frac{xy}{(x^2 + y^2)^{\frac{1}{3}}}$ if $(x,y)$ are different from $(0,0)$ and $0$ if $(x,y)= (0,0)$ and we want to check if $f(x,y)$ is differential at $(0,0)$. I have proved that f is continuous at $(0,0)$ and calculated the partial derivatives. Also,using limits i have shown that $\frac{df}{dx} = 0$ if $(x,y) = (0,0)$ and $\frac{df}{dy} = 0$ if $(x,y)=(0,0)$. All we need to do now is prove that they are continuous at $(0,0)$. Although i've calculated the partial derivatives,i'm having trouble at bounding them to prove the continuity. Any help would be appreciated.

Answer 1

Since the partial derivatives at $(0,0)$ are $0$, the function is differentiable at $(0,0)$ if and only if $D_{(0,0)}f$ is the null function, which means that$$\lim_{(x,y)\to(0,0)}\frac{xy}{(x^2+y^2)^{5/6}}=0.\tag1$$This is true: if $\|(x,y)\|=r$, then $|xy|\leqslant r^2$ and $(x^2+y^2)^{5/6}=r^{5/3}$. Therefore,$$\left|\frac{xy}{(x^2+y^2)^{5/6}}\right|\leqslant\sqrt[3]r$$and so $(1)$ holds.

Answer 2

If you switch to polar coordinates, it becomes clear $z:=f(x,y)=r^{4/3}\cos\theta\sin\theta$, where $x=r\cos\theta$ and $y=r\sin\theta$ with $r\geqslant0$ and $-\pi<\theta\leqslant\pi$. Then $\partial z/\partial\theta$ and $\partial z/\partial r$ are both bounded by $\frac43r^{1/3}$ as $r\to0$ along any path near the origin.