Show that $f(x,y)$=$\sqrt[3]{\lvert x^2-(y+1)^2 \lvert}\sin(\lvert x+y+1 \lvert))$ is differentiable at $(0,-1)$, $(1,0)$ and $(-1,0)$

Question

I want to show that this function is differentiable in these points.

At $(0,-1)$ i use the definition to check the differentiability:

At $(0,-1)$ i have that $f(0,-1)=0$, $f_y(0,-1)=0$ and $f_x(0,-1)=0$. To prove differentiability i need to check if the limit $$\lim_{(x,y)\to (0,-1)}\frac{f(x,y)-f(0,-1)-f_x(0,-1)(x-0)-f_y(0,-1)(y+1)}{\sqrt{(x-0)^2+(y+1)^2}}$$ equals to zero. This limit becomes

$$\lim_{(x,y)\to (0,-1)}\frac{\sqrt[3]{\lvert x^2-(y+1)^2 \lvert}\sin(\lvert x+y+1 \lvert))}{\sqrt{x^2+(y+1)^2}}$$

I am not sure how to proceed. Thanks for the help in advance!!

Answer 1

Hint. By the change of variables $$x=\rho \cos \theta,\quad y+1=\rho \sin \theta,$$ you may observe that $$\lim_{(x,y)\to (0,-1)}\frac{\sqrt[3]{\lvert x^2-(y+1)^2 \lvert}\sin(\lvert x+y+1 \lvert))}{\sqrt{x^2+(y+1)^2}}=\lim_{\rho\to 0}\frac{\sqrt[3]{\rho^2\lvert \cos ^2 \theta-\sin ^2 \theta \lvert}\sin(\rho\lvert \sin \theta+\cos \theta \lvert))}{\rho}=0$$ because $$\left| \frac{\sqrt[3]{\rho^2\lvert \cos ^2 \theta-\sin ^2 \theta \lvert}\sin(\rho\lvert \sin \theta+\cos \theta \lvert))}{\rho}\right|\leq \frac{\rho(\sin \theta+\cos \theta)}{\rho}\cdot \sqrt[3]{\rho^2\lvert \cos ^2 \theta-\sin ^2 \theta \lvert}=$$ $$=(\sin \theta+\cos \theta) \rho^{\frac23}\sqrt[3]{|\cos2 \theta|}\to 0$$ for $\rho\to0$.