Let $f\left(x\right)=x\left|x\right|andg\left(x\right)=\sin x ;$  
Consider the statements:
Statement 1: $gof$ is differentiable at $x=0$ and its derivative is continuous at that point.
Statement 2: $gof$ is twice differentiable at $x=0;$ then

 $f\left(x\right)=x\left|x\right|andg\left(x\right)=\sin x$
 $gof\left(x\right)=\sin \left(x\left|x\right|\right)=\{\begin{array}{cc}-\sin x^2,& x<0\\ \sin x^2,& x\ge 0\end{array}$
 $\therefore \left(gof\right)\text{'}\left(x\right)=\{\begin{array}{cc}-2x\cos x^2,& x<0\\ 2x\cos x^2,& x\ge 0\end{array}$
Clearly, L $\left(gof\right)\text{'}\left(0\right)=0=R\left(gof\right)\text{'}\left(0\right)$ 
Therefore, $gof$ is differentiable at $x=0$ and also its derivative is continuous at $x=0.$
Now, $\left(gof\right)"\left(x\right)=\{\begin{array}{cc}-2\cos x^2+4x^2\sin x^2,& x<0\\ 2\cos x^2+4x^2\sin x^2,& x\ge 0\end{array}$ 
 $\therefore L\left(gof\right)"\left(0\right)=-2andR\left(gof\right)"\left(0\right)=2$
Hence $gof"\left(0\right)doesnotexist.$