Let $f\left(x\right)=\left\{\begin{array}{c}\frac{\sin \left|x^2-5x+6\right|}{x^2-5x+6},x\ne 2,3\\ 1,x=2\text{ or }3\end{array}\right.$ the set of all points where f is differentiable is

$$\begin{gathered} x^2-5x+6<0\Rightarrow 2<x<3 \\ {x}^2-5x+6>0\Rightarrow x<2 or x>3 \\ f\left(x\right)=\left\{\begin{array}{l}\frac{\sin (x^2-5x+6)}{x^2-5x+6} if x<2\\ 1 if x=2\\ \frac{\sin (-(x^2-5x+6\left)\right)}{x^2-5x+6} if 2<x<3\\ 1 if x=3\\ \frac{\sin (x^2-5x+6)}{x^2-5x+6} if x>3\\ \end{array}\right. \\ Here L.H.L=\underset{x\to 2}{\lim }\frac{\sin (x^2-5x+6)}{x^2-5x+6}=1 \\ R.H.L=\underset{x\to 2^{+}}{\lim }\frac{\sin (-(x^2-5x+6\left)\right)}{x^2-5x+6}=-1 \\ \therefore L.H.L\ne R.H.L \\ similarly at x=3 ,L.H.L\ne R.H.L \\ \therefore f\left(x\right) is not continuous at x=2,3 \end{gathered}$$

The function is clearly differentiable except possible at $x = 2,3$