Continuity of Derivative at a point.

Yes, consider $f(x) = \begin{cases} x^2 & x \in \mathbb{Q} \\ 0 & x \in \mathbb{R} \setminus \mathbb{Q} \end{cases}$

$f$ is differentiable at $0$ and nowhere else.

This is not exactly an answer to your question, but I think the source of your confusion is that you seem to believe that the left/right hand derivatives $$f'_\pm(a)=\lim_{h\to 0^\pm} \frac{f(a+h)-f(a)}{h}$$ are the same things as the left/right hand limits of the derivative $$\lim_{h\to 0^\pm} f'(a+h).$$ They coincide in simple cases, but not in general. For example, if $$f(x)=\begin{cases}1,&x \ge 0 \\ 0,&x < 0\end{cases}$$ then $f'(x)=0$ for all $x\neq 0$, so $\lim_{x\to 0^\pm} f'(x)=0$, but $f'(0)$ doesn't exist (since $f$ is discontinuous at $x=0$). More precisely, the right hand derivative $f'_+(0)$ is zero, but the left hand derivative ${f}'_{-}(0)$ is undefined.

Yes, it's possible! Consider the function

$$f(x) = x^2 W(x)$$

where $W$ is the Weierstrass function. At $x=0$ the derivative is $0$, but if it were differentiable anywhere else then $W$ would be differentiable too.