Let $f :R\to R, g :R\to R$ and $h :R\to R$ be differentiable functions such that $f\left(x\right)=x^3+3x+2, g\left(f\left(x\right)\right)=x$ and $h\left(g\left(g\left(x\right)\right)\right)=x$ , for all $x\in R.$ Then,
 

If $f^{\prime }\left(x\right)=3x^2+3$
$\Rightarrow g\left(f\left(x\right)\right)=\mathit{x}$
$\Rightarrow g^{\prime }\left(f\left(x\right)\right)=\frac{1}{f^{\prime }\left(x\right)}$
$\mathit{Also\; f}\left(0\right)=\mathrm{2 }$   ($\mathit{Put\; x}=0$)
So, $g^{\prime }\left(2\right)=\frac{1}{f^{\prime }\left(0\right)}$ 
$\Rightarrow$ $g\prime \left(2\right)=\frac{1}{3}$
If $h(g\left(g\left(x\right)\right)=x$
$h^{\prime }\left(g\left(g\left(x\right)\right)\right)=\frac{1}{g^{\prime }\left(g\left(x\right)\right).g^{\prime }\left(x\right)}$
If, $g\left(g\left(x\right)\right)=1$  $g\left(f\left(x\right)\right)=x\Rightarrow g^{-1}\left(x\right)=f\left(x\right)$
$\Rightarrow \mathit{g}\left(x\right)=g^{-1}\left(1\right)\Rightarrow \mathit{g}\left(x\right)=f\left(1\right)=6$
To find, $g^{\text{'}}\left(6\right)$ use $x=1$ in
$g^{\text{'}}\left(\mathrm{f}\left(\mathrm{x}\right)\right)=\frac{1}{f^{\text{'}}\left(1\right)}$ as $f\left(1\right)=6$

To find, $g^{\text{'}}\left(236\right)$, use $x=6$ as
$f\left(6\right)=236$
$\therefore h^{\text{'}}\left(1\right)=\frac{1}{g^{\text{'}}\left(6\right)\mathrm{.}{g}^{\text{'}}\left(236\right)}=\frac{1}{\frac{1}{6}\mathrm{.}\frac{1}{111}}\Rightarrow h^{\text{'}}\left(1\right)=666$
If $g\left(g\left(x\right)\right)=0$
$\therefore g\left(x\right)=g^{-1}\left(0\right) \Rightarrow g\left(x\right)=f\left(0\right)$

$\Rightarrow g\left(x\right)=2\Rightarrow x=g^{-1}\left(2\right)$

$\Rightarrow x=f\left(2\right) \Rightarrow x=16$
$\therefore h\left(0\right)=16$
$g\left(x\right)=3 \Rightarrow x=g^{-1}\left(3\right)$

$\Rightarrow x=f\left(3\right) \Rightarrow x=38$
So $h\left(g\left(3\right)\right)=38$