The function f:ℝ→ ℝ defined by f(x)=|x|^3 is

$\lim_{h \to 0}f(h)-f(0)/h=\lim_{h \to 0}h|h|=0$. Hence $f$ is differentiable at $x=0$. $f'(x)=\begin{cases} 3x^2& \text{if}\hspace{.2cm} x >0 \\ 0 & \text{if}\hspace{.2cm} x=0\\ -3x^2,& \text{if}\hspace{.2cm} x>0\\ \end{cases}$. From here we can calculate $f''(0)=0. So $f''(x)=\begin{cases} 6x& \text{if}\hspace{.2cm} x >0 \\ 0 & \text{if}\hspace{.2cm} x=0\\ -6x,& \text{if}\hspace{.2cm} x>0\\ \end{cases}$. This function is not differentiable at $x=0$.