$f$ differentiable except in $1$ point
Yes, $f$ is differentiable at $0$.
Hint. By the Mean Value Theorem used in the interval $[0,x]$ for all $x>0$, $$\frac{f(x)-f(0)}{x-0}=f'(t_x)$$ for some $t_x\in (0,x).$ Hence $$\lim_{x\to 0^+}\frac{f(x)-f(0)}{x-0}=\lim_{x\to 0^+}f'(t_x)=\lim_{t\to 0^+}f'(t)$$ The same can be done in $[x,0]$ fo $x<0$.
f is diferenciable at 0. Hint: Use the definition of the derivative at 0, and by continuity of f at 0 numerator converges to 0 and denominator converges to 0 then apply L'hopital's Rule.