A simple substitution for $x$ . Isn't it?

$\begin{aligned} f(x) & = \ln \left(\frac {1+x}{1-x} \right) \\ \implies f(g(x)) & = \ln \left(\frac {1+g(x)}{1-g(x)} \right) \\ & = \ln \left(\frac {1+\frac {3x+x^3}{1+3x^2}}{1-\frac {3x+x^3}{1+3x^2}} \right) \\ & = \ln \left(\frac {1+3x^2+3x+x^3}{1+3x^2-3x-x^3}\right) \\ & = \ln \left(\frac {(1+x)^3}{(1-x)^3}\right) \\ & = \ln \left(\frac {1+x}{1-x}\right)^3 \\ & = 3 \ln \left(\frac {1+x}{1-x}\right) \\ & = \boxed{3f(x)} \end{aligned}$

Note: This may/may not be treated as a proper solution. It is more of a hint.

An intuitive way to approach this problem would be to directly replace $x$ with $g(x)$ in $f(x)$ to get

$f \circ g(x) = \ln \left( \dfrac{1+g(x)}{1-g(x)} \right)$

which upon some algebraic manipulations gives

$f \circ g(x) = \ln {\left( \dfrac{1+x}{1-x} \right)}^3 = 3 \ln \left( \dfrac{1+x}{1-x} \right) = 3f(x)$

Another way to approach this is by hyperbolic trigonometry and the inverse hyperbolic trigonometry . Ponder over how are these functions related to them.