If g ( x ) = 1 6 f 3 x 2 - 1 + 1 2 f 1 - x 2 , ∀ x ∈ R , where f '' ( x ) > 0 ,…

$g\left(x\right)=\frac{1}{6}f\left(3x^2-1\right)+\frac{1}{2}f\left(1-x^2\right)$

$\Rightarrow g^{\text{'}}\left(x\right)=\frac{1}{6}{f}^{\text{'}}\left(3x^2-1\right)·6x+\frac{1}{2}{f}^{\text{'}}\left(1-x^2\right)·\left(-2x\right)$

$\Rightarrow g^{\text{'}}\left(x\right)=x\left[f^{\text{'}}\left(3x^2-1\right)-f^{\text{'}}\left(1-x^2\right)\right]$

For $g\left(x\right)$ to be increasing function

$\Rightarrow g^{\text{'}}\left(x\right)=x\left[f^{\text{'}}\left(3x^2-1\right)-f^{\text{'}}\left(1-x^2\right)\right]>0$

$\text{So,if} x>0$

$\Rightarrow \left[f^{\text{'}}\left(3x^2-1\right)-f^{\text{'}}\left(1-x^2\right)\right]>0$

$\Rightarrow f^{\text{'}}\left(3x^2-1\right)>f^{\text{'}}\left(1-x^2\right)$

$\Rightarrow \left(3x^2-1\right)>\left(1-x^2\right)$

$\Rightarrow 4 x^2>2$

$\Rightarrow x^2 >\frac{1}{2}\Rightarrow x\in \left(\sqrt{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right. } ,\infty \right)$

$\text{if} x<0$

$\Rightarrow f^{\text{'}}\left(3x^2-1\right)<f^{\text{'}}\left(1-x^2\right)$

$\Rightarrow \left(3x^2-1\right)<\left(1-x^2\right)$

$\Rightarrow x^2 <\frac{1}{2} \& x<0\Rightarrow x\in \left(-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right. ,0\right)$

$\Rightarrow g\left(x\right) \text{increasing in }\left(-\sqrt{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} ,0\right)\cup \left(\sqrt{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} ,\infty \right)$