Answer
$$\frac{{dy}}{{dx}} = 24{\left[ {\sin 3x + \cot \left( {{x^3}} \right)} \right]^7}\left( {\cos 3x - {x^2}{{\csc }^2}{x^3}} \right)$$
Work Step by Step
$$\eqalign{
& y = {\left[ {\sin 3x + \cot \left( {{x^3}} \right)} \right]^8} \cr
& {\text{differentiate with respect to }}x \cr
& \frac{{dy}}{{dx}} = {D_x}{\left[ {\sin 3x + \cot \left( {{x^3}} \right)} \right]^8} \cr
& {\text{use the chain rule}} \cr
& \frac{{dy}}{{dx}} = 8{\left[ {\sin 3x + \cot \left( {{x^3}} \right)} \right]^7}{D_x}\left[ {\sin 3x + \cot \left( {{x^3}} \right)} \right] \cr
& {\text{use the differentiation rule for trigonometric functions}}{\text{, then}} \cr
& \frac{{dy}}{{dx}} = 8{\left[ {\sin 3x + \cot \left( {{x^3}} \right)} \right]^7}\left( {3\cos 3x - {{\csc }^2}{x^3}\left( {3{x^2}} \right)} \right) \cr
& {\text{simplifying}} \cr
& \frac{{dy}}{{dx}} = 24{\left[ {\sin 3x + \cot \left( {{x^3}} \right)} \right]^7}\left( {\cos 3x - {x^2}{{\csc }^2}{x^3}} \right) \cr} $$