$f\left(x\right)=\mathrm{sgn}\left(x\right)=\left\{\begin{array}{c}-1, x<0 \\ 0, x=0\\ 1, x>0\end{array}\right\}$

$g\left(x\right)=x\left(x^2-5x+6\right)=x\left(x-2\right)\left(x-3\right)$

$f\left(g\left(x\right)\right)=\mathrm{sgn}\left(g\left(x\right)\right)=\mathrm{sgn}\left(x\left(x-2\right)\left(x-3\right)\right)=\left\{\begin{array}{c}-1 , x\left(x-2\right)\left(x-3\right)<0\\ 0 , x\left(x-2\right)\left(x-3\right)=0\\ 1 , x\left(x-2\right)\left(x-3\right)>0\end{array}\right\}$

Let $h\left(x\right)=f\left(g\left(x\right)\right)$

$h\left(x\right)=-1$if $x\in \left(-\infty ,0\right)\cup \left(2,3\right)$

$h\left(x\right)=0,$ if $x=0,2,3$

$h\left(x\right)=1$if $x\in \left(0,2\right)\cup \left(3,\infty \right)$

Therefore, $f\left(g\left(x\right)\right)$ is discontinuous at three points $0, 2, 3$.