Checking the Monotonic function 
Let $g''(x)>0$, such that $x \in \mathbb{R} $ and $f(x)=2g(2x^3–3x^2)+g(6x^2–4x^3–3)$, $x \in \mathbb{R} $,then
(A)in interval $x \in (1,\infty) $; $f(x)$ is increasing function
(B)in interval $x \in (-\infty , \frac{-1}{2}) $; $f(x)$ is decreasing function
(C) in interval $x \in (0,\infty) $; $f(x)$ is monotonic function
(D)in interval $x \in (-\infty , 0) $; $f(x)$ is non-monotonic function

My approach is as follow
as $g''(x)>0$ therefore $g'(x)$ is an increasing function.
But while substituting I end up getting $f'(x)=-3g'(x)$

I will just give a hint.
$g''(x)>0$ implies that $g'(x)$ is increasing, meaning $g'(a)>g'(b)$ whenever $b<a$.
Now
$$
\begin{split}
f'(x)&= 2g'(2x^3-3x^2)\cdot (6x^2-6x) + g'(6x^2-4x^3-3) \cdot (12x - 12x^2) \\
&= 12x(x^2-1)g'(2x^3-3x^2) + 12x(1-x^2)g'(6x^2-4x^3-3) \\
&= 12x(x^2-1)g'(2x^3-3x^2) - 12x(x^2-1)g'(6x^2-4x^3-3) \\
&= 12x(x^2-1)\big(g'(2x^3-3x^2) - g'(6x^2-4x^3-3)\big)
\end{split}
$$
Now $12x(x^2-1)>0$ if $x>1$. But what can you say about the other term, $g'(2x^3-3x^2) - g'(6x^2-4x^3-3)$? Maybe examine the functions on the inside....perhaps plotting them on the same axes will help you (given my first comment).
