Given $f(x)=\left | x-1 \right |+\left | x+2 \right |+\left | x-3 \right |$
$\therefore f(x)=(1-x)+-(x-2)+(3-x)$ For $x<-2$
$\quad \quad =2-3x$ For $x<-2$
$f(x)=(1-x)+(x+2)+(3-x)$ For $-2\leq x< 1$
$\quad \quad=6-x$ For $-2\leq x< 1$
$f(x)=x-1+x+2+3-x$ For $1\leq x< 3$
$\quad \quad =4+x$ For $1\leq x< 3$
$f(x)=x-1+x+2+x-3$
$\quad \quad =3x-2$ For $x\geq 3$
$\therefore f(x) =\left \{ 2-3x \right.\quad x<-2$
$ \left \{ 6-x \right.\quad -2\leq x< 1$
$ \left \{ 4+x \right.\quad 1\leq x< 3 $
$ \left \{ 3x-2 \right.\quad x\geq 3$
$ {f}'(x)=\left { -3 \right. \quad x<-2 $
$ \left { -1 \right. \quad -2\leq x< 1 $
$ \left { 1 \right. \quad \ 1\leq x< 3 $
$ \left { 3 \right.\quad x\geq 3$
$\therefore f(x)$ is not differentiable at x=-2,1,3