Answer
$$\frac{{18x{{\ln }^2}\left( {3{x^2} + 2} \right)}}{{3{x^2} + 2}}$$
Work Step by Step
$$\eqalign{
& \frac{d}{{dx}}\left( {{{\ln }^3}\left( {3{x^2} + 2} \right)} \right) \cr
& {\text{use the chain rule}} \cr
& = 3{\ln ^2}\left( {3{x^2} + 2} \right)\frac{d}{{dx}}\ln \left( {3{x^2} + 2} \right) \cr
& {\text{use the rule }}\frac{d}{{dx}}\left( {\ln u} \right) = \frac{1}{u}\frac{{du}}{{dx}},{\text{ letting }}u = 3{x^2} + 2 \cr
& = 3{\ln ^2}\left( {3{x^2} + 2} \right)\left( {\frac{1}{{3{x^2} + 1}}} \right)\frac{d}{{dx}}\left( {3{x^2} + 2} \right) \cr
& = 3{\ln ^2}\left( {3{x^2} + 2} \right)\left( {\frac{1}{{3{x^2} + 1}}} \right)\left( {6x} \right) \cr
& {\text{simplify}} \cr
& = \frac{{18x{{\ln }^2}\left( {3{x^2} + 2} \right)}}{{3{x^2} + 2}} \cr} $$
