$f(x)=3x-\sin(\frac{\pi x}{2})$, $\quad$ $g(x)=x^3+2x-\sin(\frac{\pi x}{2})$ 
$f:\mathbb R \rightarrow\mathbb R$ and $g:\mathbb R \rightarrow\mathbb R$ are two functions such that
$f(x)=3x-\sin(\frac{\pi x}{2})$, $\quad$ $g(x)=x^3+2x-\sin(\frac{\pi x}{2})$
then choose the correct statements

(A) $\frac{d}{dx}(f^{-1}(g^{-1}(x)))$ at $x=-12$ is $\frac{2}{3(28+\pi)}$
(B) $\frac{d}{dx}(f^{-1}(g^{-1}(x)))$ at $x=-12$ is $\frac{2}{3(28-\pi)}$
(C) Area bounded by $y=f^{-1}(x)$ and $y=g^{-1}(x)$ is $1$
(D) Area bounded by $y=f^{-1}(x)$ and $y=g^{-1}(x)$ is $\frac{1}{2}$
My Method:
$(f^{-1}(g^{-1}(x)))=(g(f(x)))^{-1}$
$\implies$
$\frac{d}{dx}(f^{-1}(g^{-1}(x)))=\frac{d}{dx}(g(f(x)))^{-1}$
$\implies$
$\frac{d}{dx}(f^{-1}(g^{-1}(x)))=\frac{-g'(f(x))f'(x)}{(g(f(x))^{2}}$
But now I'm stuck.
For Option (C)
Because intersection points are $x=-1,x=0,x=1$
So area will be $\int_{-1}^{1}|g(x)-f(x)|dx$
Solution to Option (A) given in my Solution Image is as follow:
$\frac{d}{dx}(f^{-1}(g^{-1}(x)))=\frac{1}{\frac{d}{dx}(g(f(x)))_{x=-1}}$
It seem to me that they must have used $f(g(x))=x$ if $f(x)$ and $g(x)$ are Inverse of each other. But I am not sure how did they use in above problem
Please Help me in this question.
 A: 
$$\frac{d}{dx}(f^{-1}(g^{-1}(x)))=\frac{1}{\frac{d}{dx}(g(f(x)))_{x=-1}}$$

Let $y=f^{-1}g^{-1}(x).$ Then $$gf(y)=x.$$ Differentiating with respect to $x$ gives
\begin{align}\frac{\mathrm d}{\mathrm dy}\left(gf(y)\right)\frac{\mathrm dy}{\mathrm dx}&=1
\\\frac{\mathrm d}{\mathrm dx}\left(f^{-1}g^{-1}(x)\right)&=\frac1{\frac{\mathrm d}{\mathrm dy}\left(gf(y)\right)}
\\\frac{\mathrm d}{\mathrm dx}\left(f^{-1}g^{-1}(x)\right)\bigg|_{x=-12}&=\frac1{\frac{\mathrm d}{\mathrm dy}\left(gf(y)\right)\bigg|_{y=-1}}
\\\ &=\frac1{\frac{\mathrm d}{\mathrm dx}\left(gf(x)\right) \bigg|_{x=-1}}.\end{align}

My Method:
$(f^{-1}(g^{-1}(x)))=(g(f(x)))^{-1}$
$\implies$
$\frac{d}{dx}(f^{-1}(g^{-1}(x)))=\frac{d}{dx}(g(f(x)))^{-1}$
$\implies$
$\frac{d}{dx}(f^{-1}(g^{-1}(x)))=\frac{-g'(f(x))f'(x)}{(g(f(x))^{2}}$


*

*Your main error was mixing up reciprocal and inverse.


*A minor secondary issue is the unnecessary use of the $\implies$ symbol.
Just as $=$ doesn't mean ‘outputs’,  $\implies$ doesn't mean ‘therefore’.
