Answer
$$g'\left( x \right) = - 6{\text{sec}}{{\text{h}}^2}3x\tanh 3x$$
Work Step by Step
$$\eqalign{
& g\left( x \right) = {\text{sec}}{{\text{h}}^2}3x \cr
& {\text{Differentiate}} \cr
& g'\left( x \right) = \frac{d}{{dx}}\left[ {{\text{sec}}{{\text{h}}^2}3x} \right] \cr
& {\text{By the chain rule}} \cr
& g'\left( x \right) = 2{\text{sech}}3x\frac{d}{{dx}}\left[ {{\text{sech}}3x} \right] \cr
& {\text{Using the Derivatives and Integrals of Hyperbolic Functions}} \cr
& g'\left( x \right) = 2{\text{sech}}3x\left( { - {\text{sech}}3x\tanh 3x} \right)\left( 3 \right) \cr
& g'\left( x \right) = - 6{\text{sech}}3x\left( {{\text{sech}}3x\tanh 3x} \right) \cr
& g'\left( x \right) = - 6{\text{sec}}{{\text{h}}^2}3x\tanh 3x \cr} $$
