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

Question

I want to show that this function is differentiable at these points. At $(0,-1)$ i use the definition to check the differentiability: $\lim_{\substack{x\to 0 \\ y\to -1}} \frac{\sqrt[3]{\lvert x^2-(y+1)^2 \lvert}\sin(\lvert x+y+1 \lvert))}{\sqrt{x^2+(y+1)^2}}$.

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

Answer 1

Hint at $(0,-1)$: Note $|x^2-(y+1)^2|\le x^2+(y+1)^2$ and $|\sin |x+y+1||\le |x+y+1|.$ So does

$$\lim_{(x,y)\to (0,-1)} \frac{(x^2+(y+1)^2)^{1/3}|x+y+1|}{(x^2+(y+1)^2)^{1/2}}=0?$$