We have,

$g\left(x\right)=\left\{\begin{array}{cc}\underset{0\le t\le x}{\max \left\{t^3-6t^2+9t-3\right\}},& 0\le x\le 3\\ 4-x,& 3<x\le 4\end{array}\right.$

Let

$f\left(x\right)=x^3-6x^2+9x-3$

$\Rightarrow f^{\text{'}}\left(x\right)=3x^2-12x+9$

$\Rightarrow f^{\text{'}}\left(x\right)=3(x-1)(x-3)$

For critical points:

$f^{\text{'}}\left(x\right)=0$

$\Rightarrow x=1, 3$

And,

$f^{"}\left(x\right)=6x-12$

Then,

$f^{"}\left(1\right)=6-12=-6<0$(maxima)

$f^{"}\left(3\right)=6>0$(minima)

Hence,

$g\left(x\right)=\left[\begin{array}{rl}f\left(x\right),& 0\le x\le 1\\ 1,& 1\le x\le 3\\ 4-x,& 3<x\le 4\end{array}\right.$

Hence, $g\left(x\right)$ is continuous function.

$g^{\text{'}}\left(x\right)=\left[\begin{array}{cc}3(x-1)(x-3),& 0<x<1\\ 0,& 1<x<3\\ -1,& 3<x<4\end{array}\right.$

Clearly, $g\left(x\right)$ is non-differentiable at $x=3$.