Let $f\left(x\right)=x\left|x\right|\text{ and }g\left(x\right)=\sin x$

Statement - 1: $gof$is differntiable at $x = 0$ and its derivative is continuous at that point.

Statement - 2 : $gof$is twice differentiable at $x = 0$

$Given f\left(x\right)=x\left|x\right|\text{ and }g\left(x\right)=\sin x$

$\Rightarrow gof\left(x\right)=\sin \left(x\right|x\left|\right)=\left\{\begin{array}{cc}-\sin x^2,& x<0\\ \sin x^2,& x\ge 0\end{array}\right.$

$\therefore \left(\text{ gof }{)}^{\mathrm{\prime }}\right(x)=\left\{\begin{array}{cc}-2x\cos x^2,& x<0\\ 2x\cos x^2,& x\ge 0\end{array}\right.$

clearly, $L\left(gof{)}^{\mathrm{\prime }}\right(0)=0=R(gof{)}^{\mathrm{\prime }}\left(0\right)$

Therefore, $gof$ is differentiable at $x=0$ also its derivative is continuous at $x=0$

$\therefore L\left(\text{ gof }{)}^{\prime \prime }\right(0)=-2\text{ and }R(gof{)}^{\prime \prime }\left(0\right)=2\text{ Hence, }\left(\text{ gof }{)}^{\prime \prime }\right(2)$ does not exist.