If $g\left( x \right) =2f\left( 2{ x }^{ 3 }-3{ x }^{ 2 } \right) +f\left( 6{ x }^{ 2 }-4{ x }^{ 3 }-3 \right)$ $\forall \ x\ \in \ R$ and $f^{''}\left( x \right) > 0$, $\forall \ x \in \ R$, then $g\left ( x \right)$ is increasing in the interval

Any function $f(x)$ is Increasing

$\text { if } f^{\prime}(x)>0$

decreasing if $f^{\prime}(x)<0$

Here,

$\begin{array}{l} g(x)=2 f\left(2 x^{3}-3 x^{2}\right)+f\left(6 x^{2}-4 x^{3}-3\right) \\ g^{\prime}(x)=2 f^{\prime}\left(2 x^{3}-3 x^{2}\right)\left[6 x^{2}-6 x\right]+f^{\prime}\left(6 x^{2}-4 x-3\right)[12 x-12 x] \\ \Rightarrow\left(12 x^{2}-12 x\right) f^{\prime}\left(2 x^{3}-3 x^{2}\right)+\left(12 x+2 x^{2}\right) f^{\prime}\left(6 x^{2}-4 x^{3}-3\right) \\ \Rightarrow 12 x^{2}-12 x\left[f^{\prime}\left(2 x^{3}-3 x^{2}\right)-f^{\prime}\left(6 x^{2}-4 x^{3}-3\right)\right] \\ \left.2 x^{3}-3 x^{2}\right\rangle 6 x^{2}-4 x^{3}-3 \\ \end{array}$

if $f^{\prime \prime}(x)>0$

$f^{\prime}(x)$ increasing

$\begin{array}{l} 2 x^{3}-3 x^{2}+1>0 \\ (x-1)(x-1)(2 x+1)>0 \end{array}$

$f^{\prime}\left(x_{1}\right)>f^{\prime}\left(x_{2}\right)$

$x_{1}>x_{2}$

if

$\begin{array}{l} x>-\frac{1}{2} \\ \Rightarrow 2 x^{3}-3 x^{2}>6 x^{2}-4 x^{3}-3 \\ f^{\prime}\left(2 x^{3}-3 x^{2}\right)>f^{\prime}\left(6 x^{2}-4 x^{3}-3\right) \\ g^{\prime}(x)=12 x(x-1)\left[f^{\prime}\left(2 x^{3}-3 x^{2}\right)-f^{\prime}\left(6 x^{2}-4 x^{3}-3\right)\right] \\ \end{array}$

$x \in\left(-\frac{1}{2}, 0\right) \cup(1, \infty)$