Let $f(x)=x|x|$ and $g(x)=\sin x$

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

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

Statement-1 is True, Statement-2 is True; statement-2 is a correct explanation for Statement-1.

$g(f(x)) =\begin{cases}-\sin x^2, & x<0 \\ \sin x^2, & x \geq 0\end{cases}$

$LHL]_{x=0^-}=0$

$RHL]_{x=0^+}=0$

$[g(f(x))]'=\begin{cases}-2x\cos x^2, & x<0 \\ 2x\cos x^2, & x \geq 0\end{cases}$

$LHD]_{x=0^-}=0$

$RHD]_{x=0^+}=0$

$LHD=RHD→gof(x)$ differentiable at $x = 0$ and continuous derivable at $x = 0$

RHD of $(g(f(0)))'=\lim\limits_{x \rightarrow 0^{+}}\frac{2h\cos h^2-gof'(0)}{h}=2$

LHS of $(g(f(0)))'=\lim\limits_{x \rightarrow 0^{-}}\frac{-2h\cos h^2-gof'(0)}{-h}=2$

$gof$ is twice differentiable at $x = 0$