Let $f(x)=|x|x$ then $f''(0)$ does not exist.
Why?
If $x>0$, $f'(x)=2x$ and if $x<0$, $f'(x)=-2x$.
Then when $x=0$, does $f'(x)$ also not exist?
Let $f(x)=|x|x$ then $f''(0)$ does not exist.
Why?
If $x>0$, $f'(x)=2x$ and if $x<0$, $f'(x)=-2x$.
Then when $x=0$, does $f'(x)$ also not exist?
Write $f$ as: $ f(x) = \begin{cases} x^2 & \text{if } x \ge 0 \\ -x^2 & \text{if } x < 0\end{cases} $
It's easy to see that $f$ is differentiable everywhere. The case $x = 0$ gives $f'(0) = 0$ for the limits from both sides.
We have: $ f'(x) = \begin{cases} 2x & \text{if } x \ge 0 \\ -2x & \text{if } x < 0\end{cases} $
In other words, $f'(x) = 2|x|$, which is not differentiable at $x = 0$.
$f'(x)= 2x$ if $x\geq0$ and $f'(x)=-2x$ if $x\leq 0$. Note that $f'$ is well defined (because $2\times 0 = -2 \times 0$.
Now for $f''$ we have $f''(x)= 2$ if $x\geq 0$ and $f''(x)= -2$ if $x\leq 0$. This isn't well defined in 0.