# Differentiable but not continuously-differentiable function: not the usual one

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.

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