We have

$f\left(x\right)=\left\{\begin{array}{l}-x,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}-1\le x<0\\ 1-x,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\le x<1\\ x,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\le x<2\\ 1+x,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}2\le x<3\\ 5,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}x=3\end{array}\right.$

Clearly,$f \left(x\right)$ is discontinuous at $x = 0,1, 2, and 3.$

So, it is not differentiable at these points.

At $x=-1$, we have

$\underset{x\to -1^{+}}{lim} f\left(x\right)=\underset{x\to -1^{+}}{lim} -x=1=f(-1)$

So, it is continuous at $x=-1$.

Also, (RHD at $x=-1)=-1$ (a finite number). 

Therefore, $f \left(x\right)$ is differentiable at$x=-1.$