Consider the function
$$f(x,y) =(x^2 + y^2)\sin\left(\frac{1}{\sqrt{x^2 + y^2}}\right)$$
The partial derivative with respect to $x$ are equal to
$$\frac{\partial}{\partial x}f(x,y) =\left\{\begin{array} \\ 0 &\text{ if $x = 0, y = 0$ } \\ 2x \sin(1/|x|) - sign(x)\cos(1/|x|) & \text{ if $x \ne 0, y = 0$} \\ \cos \left( \frac{1}{\sqrt{x^2 + y^2}}\right)\frac{x}{\sqrt{x^2 + y^2}} &\text{ if $x \ne 0,y \ne 0$} \end{array} \right\}$$ $$\frac{\partial}{\partial y}f(x,y) =\left\{\begin{array} \\ 0 &\text{ if $x = 0, y = 0$ } \\ 2y \sin(1/|y|) - sign(y)\cos(1/|y|) & \text{ if $x = 0, y \ne 0$} \\ \cos \left( \frac{1}{\sqrt{x^2 + y^2}}\right)\frac{y}{\sqrt{x^2 + y^2}} &\text{ if $x \ne 0,y \ne 0$} \end{array} \right\}$$
I've heard that $f(x,y)$ is differentiable even though the partial derivatives are not continuous.
However, the sources I've seen don't prove that $f(x,y)$ is differentiable. In fact, they just point out (very misleadingly) that the partial derivatives are equal to zero at $(0,0)$, which is not enough for a function to be differentiable at $(0,0)$.
Question How do I prove that $f(x,y)$ is differentiable at $(0,0)$?
The only point of contention is $(x,y) = (0,0)$.
We have $| f(x,y) - f(0,0) - 0 (x,y)| \le \|(x,y)\|^2$, hence the Frechet derivative is zero (the zero linear mapping).