Answer
The function $ f(x)=|x^2-1|$ is not differentailble at $ x=\pm1$.
Work Step by Step
Since $$|x^2-1|=0\Longrightarrow x^2-1=0\Longrightarrow (x+1)(x-1)=0.$$
The function $ f(x)=|x^2-1|$ has corners at $ x=\pm 1$, hence it is not differentailble at $ x=\pm1$.
