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**Laurent** The function $\displaystyle f:x\mapsto x^2\sin\frac{1}{x}$ is continuous at 0, $\displaystyle f'$ is not continuous at 0 (as you can check by computation). However, $\displaystyle f$ is differentiable at 0 ($\displaystyle \frac{f(x)}{x}\to 0$ as $\displaystyle x$ tends to 0)...

Your argument could be made rigorous using additional properties. You can find two directions ($\displaystyle x\mapsto(\pm x,0)$ for $\displaystyle x>0$) where different finite limits *exist* for $\displaystyle \frac{\partial f}{\partial x}$. And then use the following fact: if a continuous function is differentiable on $\displaystyle (a,b)$ and such that its derivative admits a finite limit $\displaystyle \ell$ at $\displaystyle b$, then the function is differentiable at $\displaystyle b$ and its derivative at $\displaystyle b$ is $\displaystyle \ell$. But this is a bit overcomplicated.