Let $\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}\left|\mathrm{x}\right|$ and $\mathrm{g}\left(\mathrm{x}\right)=\mathrm{sinx}$
Statement-I: gof is differentiable at x=0 and its derivative is continuous at that point.
Statement-2: gof is twice differentiable at x=0.
 

$\left(\mathrm{gof}\right)\left(\mathrm{x}\right)=\mathrm{g}\left[\mathrm{f}\left(\mathrm{x}\right)\right]=\sin \left(\mathrm{x}\left|\mathrm{x}\right|\right)=\{\begin{array}{l}-\sin {\mathrm{x}}^2,\mathrm{x}<0\\ \sin {\mathrm{x}}^2,\mathrm{x}\ge 0\end{array}$
Let the composite function  $\left(\mathrm{gof}\right)\left(\mathrm{x}\right)$ be denoted by $\mathrm{H}\left(\mathrm{x}\right).$then $\mathrm{H}\left(\mathrm{x}\right)=\{\begin{array}{l}-\sin {\mathrm{x}}^2,\mathrm{x}<0\\ \sin {\mathrm{x}}^2,\mathrm{x}\ge 0\end{array}$
${\mathrm{H}}^{\text{'}}(0-){=}_{\mathrm{x}\to 0-}^{\mathrm{Lt}}\frac{\mathrm{H}\left(\mathrm{x}\right)-\mathrm{H}\left(0\right)}{\mathrm{x}-0}{=}_{\mathrm{x}\to 0-}^{\mathrm{Lt}}\frac{-{\mathrm{sinx}}^2-0}{\mathrm{x}}=0$
$\mathrm{H}\text{'}(0+){=}_{\mathrm{x}\to 0+}^{\mathrm{Lt}}\frac{\mathrm{H}\left(\mathrm{x}\right)-\mathrm{H}\left(0\right)}{\mathrm{x}-0}{=}_{\mathrm{x}\to 0+}^{\mathrm{Lt}}\frac{\sin {\mathrm{x}}^2-0}{\mathrm{x}}=0.$
$\therefore \mathrm{H}\left(\mathrm{x}\right)$ differentiable at 0 and $\mathrm{H}\text{'}\left(0\right)=0$
Also $\mathrm{H}\text{'}\left(\mathrm{x}\right)=\{\begin{array}{l}-2\mathrm{x}\cos {\mathrm{x}}^2,\mathrm{x}<0\\ 0,\mathrm{x}=0\\ 2\mathrm{x}\cos {\mathrm{x}}^2,\mathrm{x}>0\end{array}$
Again $\mathrm{H}"\left(\mathrm{x}\right)=\{\begin{array}{l}-2{\mathrm{cosx}}^2+4{\mathrm{x}}^2{\mathrm{sinx}}^2,\mathrm{x}<0\\ 2{\mathrm{cosx}}^2-4{\mathrm{x}}^2{\mathrm{sinx}}^2,\mathrm{x}\ge 0\end{array}$
And $\mathrm{H}"(0-)=-2\mathrm{\hspace{0.17em}}$and $\mathrm{H}"(0+)=2$
Thus H(x) is not twice differentiable at x=0.