Answer
$$\frac{{dy}}{{dx}} = 6{x^2}\cosh 3x\sinh 3x + 2x{\cosh ^2}3x$$
Work Step by Step
$$\eqalign{
& y = {x^2}{\cosh ^2}3x \cr
& {\text{computing }}dy/dx \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left( {{x^2}{{\cosh }^2}3x} \right) \cr
& {\text{by the product rule}} \cr
& \frac{{dy}}{{dx}} = {x^2}\frac{d}{{dx}}\left( {{{\cosh }^2}3x} \right) + {\cosh ^2}3x\frac{d}{{dx}}\left( {{x^2}} \right) \cr
& {\text{by the chain rule}} \cr
& \frac{{dy}}{{dx}} = {x^2}\left( {2\cosh 3x} \right)\frac{d}{{dx}}\left( {\cosh 3x} \right) + {\cosh ^2}3x\frac{d}{{dx}}\left( {{x^2}} \right) \cr
& {\text{using basic formulas for differentiation}} \cr
& \frac{{dy}}{{dx}} = {x^2}\left( {2\cosh 3x} \right)\left( {\sinh 3x} \right)\left( 3 \right) + {\cosh ^2}3x\left( {2x} \right) \cr
& {\text{multiplying}} \cr
& \frac{{dy}}{{dx}} = 6{x^2}\cosh 3x\sinh 3x + 2x{\cosh ^2}3x \cr} $$