Answer
$g'\left( x \right) = \frac{{3{x^2}}}{{1 - {x^6}}}$
Work Step by Step
$$\eqalign{
& g\left( x \right) = {\tanh ^{ - 1}}\left( {{x^3}} \right) \cr
& {\text{Differentiate}} \cr
& g'\left( x \right) = \frac{d}{{dx}}\left[ {{{\tanh }^{ - 1}}\left( {{x^3}} \right)} \right] \cr
& {\text{Use Derivatives of Inverse Hyperbolic Functions }} \cr
& \frac{d}{{dx}}\left[ {{{\tanh }^{ - 1}}u} \right] = \frac{1}{{1 - {u^2}}}\frac{{du}}{{dx}},{\text{ let }}u = {x^3},{\text{ so}} \cr
& g'\left( x \right) = \frac{1}{{1 - {{\left( {{x^3}} \right)}^2}}}\frac{d}{{dx}}\left[ {{x^3}} \right] \cr
& {\text{Compute the derivative and simplify}} \cr
& g'\left( x \right) = \frac{1}{{1 - {{\left( {{x^3}} \right)}^2}}}\left( {3{x^2}} \right) \cr
& g'\left( x \right) = \frac{{3{x^2}}}{{1 - {x^6}}} \cr} $$
