Give an example, with proof, of a function differentiable at a point, whose derivative is non-differentiable that point.

The absolute value function is continuous so has an antiderivative. The antiderivative is differentiable at 0, but its derivative (the absolute value function) is not.

One of the simplest is $f(x) = \begin{cases} \phantom{+}x^2 & \text{if } x\ge0, \\ -x^2 & \text{if } x<0. \end{cases}$

The derivative is $f'(x) = \begin{cases} \phantom{-} 2x & \text{if } x>0, \\ -2x & \text{if } x<0, \\ \text{what?} & \text{if } x=0. \end{cases}$

At $x=0$ one can compute $\displaystyle f'(x)= \lim_{h\to0} \frac{f(x+h) - f(x)} h = \lim_{h\to0} \frac{\pm h^2 - 0} h =0.$

Hint:

Consider the function

$$f(x) = \begin{cases} x^3\sin\left(\frac{1}{x}\right), & \text{ if } x \neq 0 \\ 0, & \text{ if } x = 0 \end{cases} \tag{1}\label{eq1A}$$

at $x = 0$.