$\lim\limits_{x\to 0^-}f(x)=\lim\limits_{x\to 1}(x-1)=0$
$\lim\limits_{x\to 0^+}f(x)=\lim\limits_{x\to 1}(x^3-1)=0$
also $f(1)=1-1=0$
$\therefore f$ is continuous at $x=1$
Clearly $Lf'(1)=2$ and $Rf'(1)=3$
$\therefore f(x)$ is not differentiable at $x=1$
Hence (c) is the correct answer.