real analysis - Differentiable, but not continuous

Let $f:\mathbb R^2 \to\mathbb R$ be defined by

$$f(x,y) = \begin{cases}(x^2 + y^2)\sin(\frac{1}{\sqrt{x^2 + y^2}}) &\mbox{when } (x, y) \ne (0, 0) \\ 0 &\mbox{when } (x, y) = (0, 0) \end{cases}$$
Show that $f$ is differentiable at $(0,0)$ but the ith partial derivative of $f$ is not continuous at $(0,0)$

My guess is that the derivative at $(0,0)$ will be $0$ since the function is $0$ at the point. However, I am facing some trouble proving it with the help of the definition. And I am also confused about what to do once I have proven it using the definition.

Answer

The differential at origin is $Df(0,0)=(0,0)$ (not $0\in\Bbb R$) iff the following limit is zero:
$$\lim_{(x,y)\to(0,0)}{f(x,y)-f(0,0)-Df(0,0)(x,y)\over||(x,y)||}= \lim_{(x,y)\to(0,0)}{(x^2 + y^2)\sin(1/(\sqrt{x^2 + y^2})) \over||(x,y)||}=$$
$$\lim_{(x,y)\to(0,0)}\sqrt{x^2 + y^2}\sin(1/(\sqrt{x^2 + y^2}))=0.$$
But for $(x,y)\ne(0,0)$
$${\partial f\over\partial x}(x,y)=2\,x\,\sin \left({{1}\over{\sqrt{x^2+y^2}}}\right)-{{x\,\cos \left( {{1}\over{\sqrt{x^2+y^2}}}\right)}\over{\sqrt{x^2+y^2}}}$$
and ${\partial f\over\partial x}$ is discontinuous at (0,0) because
$${\partial f\over\partial x}(0,0)=0\ne \lim_{x\to 0}{\partial f\over\partial x}(x,0)= \lim_{x\to 0}2\,x\,\sin\left({1\over|x|}\right)-{{x\,\cos \left( {{1}\over |x|}\right)}\over|x|}=0-\not\exists=\not\exists.$$
(the 2D limit can't $\exists$ is some directional limit $\not\exists$)