A function $g\left(x\right)$  is defined as   $g\left(x\right)=\frac{1}{4}f(2x^2-1)+\frac{1}{2}f(1-x^2)$   and  $f^{\mathrm{\prime }}\left(x\right)$  is an increasing function.
Then $g\left(x\right)$  is increasing in the interal

$g^{\mathrm{\prime }}\left(x\right)=x{f}^{\mathrm{\prime }}(2x^2-1)-x{f}^{\mathrm{\prime }}(1-x^2)=x(f^{\mathrm{\prime }}(2x^2-1)-f^{\mathrm{\prime }}(1-x^2))$

$\Rightarrow g^{\mathrm{\prime }}\left(x\right)>0 \text{if }x>0,2x^2-1>1-x^2$  

(as $f\text{'}$  is an increasing function) 

$\Rightarrow$$3x^2>2 \Rightarrow x\in (-\mathrm{\infty },-\sqrt{\frac{2}{3}})\cup (\sqrt{\frac{2}{3}},\mathrm{\infty })$ 

$\Rightarrow$ $$\begin{gathered} x\in (\sqrt{\frac{2}{3}},\mathrm{\infty }) \\ \Rightarrow \text{if }x<0,2x^2-1<1-x^2 \end{gathered}$$

$\Rightarrow$  $3x^2<2$$\Rightarrow$  $x\in (-\sqrt{\frac{2}{3}},\sqrt{\frac{2}{3}})$ $\Rightarrow$  $x\in (-\sqrt{\frac{2}{3}},0)$