$=\left|2x+1\right|-3\left|x+2\right|+\left|x+2\right|\left|x-1\right|$

$=\left|2x+1\right|+\left|x+2\right|\left(\left|x-1\right|-3\right)$

Critical points are $x=\frac{-1}{2},-2,-1$

but $x=-2$ is making a zero twice in product so, points of non differentiability are $x=\frac{-1}{2}$ and $x=-1$

$\therefore$ Number of points of non-differentiability $=2$