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It is well-known that the function $f:\mathbb R^2\to \mathbb R $ defined by $$f(x,y)=\begin{cases}(x^2+y^2)\sin\left(\frac1 {\sqrt{x^2+y^2}}\right),&(x,y)\neq 0\\0,&(x,y)=0\end{cases}$$ is differentiable everywhere but $\dfrac{\partial f}{\partial x}(x,y)$ and $\dfrac{\partial f}{\partial y}(x,y)$ are not continuous at $(0,0)$, this is the standard example to prove that there exist differentiable but not continuously-differentiable functions (e.g., see https://math.stackexchange.com/q/3338764).

My question: is there any other (reasonable) example (from $\mathbb R^2$ to $\mathbb R$) that differs significantly from this one? I mean: no radial simmetry and not obtained by continuous transformation from the above (and possibly avoiding the $\sin$ function) and such that the calculation can be performed by undergraduate students.

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$f(x,y)=x^2\sin(1/x^2),$ where $f(0,y)=0$ might be different.

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The function given by $$ f(x,y) = \begin{cases} y^2 \arctan\left( \dfrac{x}{y^2} \right), & y \neq 0, \\ 0, & y = 0, \end{cases} $$ is differentiable everywhere, but $\partial f/\partial x$ is discontinuous at the origin. (The other partial derivative $\partial f/\partial y$ is continuous everywhere, though.)

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