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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.

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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 $|f(x, y)|\le \|(x, y)\|^2$

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