Given:$S=\left\{1,2,3,4,5,6\right\}$

Total number of onto functions from $S$ tto $S$$=6!$

Now, for $g\left(3\right)=2g\left(1\right)$:

$\begin{array}{|cc|}\hline g\left(3\right)& g\left(1\right)\\ 2& 1\\ 4& 2\\ 6& 3\\ \hline\end{array}$ 

$\therefore \begin{array}{cc}g\left(3\right)=& 2g\left(1\right)\end{array}$can be defined in $3$ ways.

$g\left(2\right),g\left(4\right),g\left(5\right) \text{\&} g\left(6\right)$ can be anything and can be defined in $4!$ ways.

$\therefore$Number of onto functions for which $\left[g\left(3\right)=2 g\left(1\right)\right]=3.4!$

Now required probability$=\frac{3.4!}{6!}=\frac{3\times 4!}{30\times 4!}=\frac{1}{10}$