# Discontinuous partials but differentiable

I am trying to provide a counterexample to the statement that differentiability implies continuous partial derivatives. So far, I have found this function: $$f (x,y) = (x^2 + y^2) \sin\left(\frac{1}{\sqrt{x^2 + y^2}}\right)$$ for $$(x,y) \neq (0,0)$$ and $$0$$ otherwise.

If I take the partial derivative with respect to $$x$$, I get: \begin{align*} f_x = \lim\limits_{h \to 0} \frac{f(h,0) - f(0,0)}{h} = \lim\limits_{h \to 0} \frac{h^2 \sin\left(\frac{1}{\sqrt{h^2}} \right)}{h} = \lim\limits_{h \to 0} h \sin\left(\frac{1}{|\sqrt{h}|}\right) = 0. \end{align*} I get the same result for $$f_y$$.

My question involves how I demonstrate that this function is actually a counterexample. In other words, how do I show that $$f(x,y)$$ is differentiable at $$(0,0)$$ and at least one of the partials is discontinuous?

An explanation of the strategy for doing this would be more than enough.

Thanks.

Compute $$\frac{\partial f}{\partial x}(x, 0)$$ when $$x\neq 0$$. It doesn't converge to $$0$$ when $$x\to 0$$.
$$f$$ is differentiable at $$(0,0)$$ because $$|f(x, y)|\le \|(x, y)\|^2$$