$$f(x,y)=\begin{cases} (x-y)^2\sin\frac1{x-y}&x\ne y\\0&x=y\end{cases}$$ I have to show that the first-order partial derivatives exist for $f$ at each point $(x,y)\in\mathbb R^2$.
First I showed it for all $(x,x)$. There, very easily, it can be shown that the partial derivatives exist and are 0. But I could not proceed with the rest of the plane.
Hint. For a function of the form $f(x,y) = F(x-y)$, the Chain Rule implies that, if $F$ is differentiable, $\partial_x f(x,y) = F'(x-y)$ and $\partial_y f(x,y) = -F'(x-y)$.