Answer
$f''(x)=6x\:g'(x^2)+4x^3g''(x^2)$
Work Step by Step
$f(x)=x\:g(x^2)$ _____(1)
Since $g$ is twice differentiable function
Differentiating (1) with respect to $x$
$f'(x)=(x)'g(x^2)+x\left[g(x^2)\right]'$
[By chain rule $[g(x^2)]'=g'(x^2)\cdot(x^2)'=2xg'(x^2)$]
$f'(x)=g(x^2)+2x^2g'(x^2)$ ____(2)
Differentiating (2) with respect to $x$
$f''(x)=[g(x^2)]'+2[x^2g'(x^2)]'$
$f''(x)=2xg'(x^2)+2[(x^2)'g'(x^2)+x^2[g'(x^2)]']$
[By Chain rule $[g'(x^2)]'=g''(x^2)\cdot (x^2)'=2xg''(x^2)$]
$f''(x)=2xg'(x^2)+2[2xg'(x^2)+2x^3g''(x^2)]$
$f''(x)=2xg'(x^2)+4xg'(x^2)+4x^3g''(x^2)$
$f''(x)=6xg'(x^2)+4x^3g''(x^2)$
Hence $\;f''(x)=6xg'(x^2)+4x^3g''(x^2)$.
