function $\mathcal R^m\to R$: $f(x,y) = y + \cos \sqrt[3]{x^2 + y^2}, M(0,0)$
What I found out is that, obviously, $f'_x = \frac{-2x\sin\sqrt[3]{x^2 + y^2}}{3(x^2 + y^2)^\frac{2}{3}}$ and $f'_y =1- \frac{2y\sin\sqrt[3]{x^2 + y^2}}{3(x^2 + y^2)^\frac{2}{3}}$ in this point couldn't be calculated and the function itself is continious. But as I got it, I need to prove that both $f'_x$ and $f'_y$ continious at that point to prove differentiability. But I got stacked at this moment.
Note that function can be not differentiable at all, just couldn't prove that by myself either.
HINT
Note that by the definition we have that
then apply the definition to check differentiability
$$\lim_{(h,k)\rightarrow (0,0)} \frac{\| f(h,k)-f(0,0)-(f_x(0,0),f_y(0,0))\cdot (h,k)\|}{\| (h,k)\|}\stackrel{?}=0$$
that is
$$\lim_{(h,k)\rightarrow (0,0)} \frac{k + \cos \sqrt[3]{h^2 + k^2}-1-k}{\sqrt{h^2 + k^2}}\stackrel{?}=0$$