Let $g\left( x \right)$ be the inverse of a function $f\left( x \right)$ and $f'\left( x \right) = \dfrac{1}{{1 + {x^3}}}$, then $g'\left( x \right)$ is equal to
1. $\dfrac{1}{{\left( {1 + {{\left\{ {f\left( x \right)} \right\}}^3}} \right)}}$
2. $\left( {1 + {{\left\{ {f\left( x \right)} \right\}}^3}} \right)$
3. $\dfrac{1}{{\left( {1 + {{\left\{ {g\left( x \right)} \right\}}^3}} \right)}}$
4. $\left( {1 + {{\left\{ {g\left( x \right)} \right\}}^3}} \right)$

Hint:An inverse function is one that reverses the operation of another function. Using given condition write the composite function $gof$, differentiate $gof$ with respect to $x$. Solve further and put the values from the given condition.

Formula Used:
Chain rule –
$\dfrac{d}{{dx}}g\left( {f\left( x \right)} \right) = g'\left( {f\left( x \right)} \right) \times \dfrac{d}{{dx}}\left( {f\left( x \right)} \right)$

Complete step by step Solution:
Given that,
$g\left( x \right)$ is the inverse of a function $f\left( x \right)$
$f\left( x \right) = y \Rightarrow g\left( y \right) = x$
$\therefore g\left( {f\left( x \right)} \right) = x - - - - - (1)$
Differentiate equation (1) with respect to $x$
$g'\left( {f\left( x \right)} \right) \times f'\left( x \right) = 1$
$g'\left( {f\left( x \right)} \right) = \dfrac{1}{{f'\left( x \right)}}$
$g'\left( y \right) = \dfrac{1}{{\left( {\dfrac{1}{{1 + {x^3}}}} \right)}}$
$g'\left( y \right) = 1 + {x^3}$
$g'\left( y \right) = 1 + {\left\{ {g\left( y \right)} \right\}^3}$
$ \Rightarrow g'\left( x \right) = 1 + {\left\{ {g\left( x \right)} \right\}^3}$

Hence, the correct option is 4.

Note: To solve this type of question, one must remember the inverse of the function. Basically, the inverse of the function means applying $f$ and then $g$ is equivalent to doing nothing. As they allow mathematical processes to be reversed, inverse methods are essential for solving equations.