$f\left(x\right)=\sqrt{x^2-3x+9}$ then the derivative of  $f\left(x\right)$ in the interval  $(0,5)$ is

$f\left(x\right)=\sqrt{x^2-3x+9}$

$=\sqrt{{(x-3)}^2}$

$f\left(x\right)=|x-3|$

$f\left(x\right)=\{\begin{array}{l}x-3,x\ge 3\\ 3-x,x<3\end{array}$

$\Rightarrow f^{|}\left(x\right)=\{\begin{array}{cc}1,& ifx\ge 3\\ -1,& ifx<3\end{array}$

$f\left(x\right)$ is differentiable at all points on the interval $(0,5)$ except at  $x=3$

$\therefore$ the derivative of $f\left(x\right)$ on the interval $(0,5)$ does not exist.