Answer
$y'=3sin(3x)cos^2(3x)(3+cos^3(3x))^{-4/3}$
Work Step by Step
Take the derivative of the equation using Power Rule, Chain Rule, and Trigonometric derivative:
$y'=\frac{-1}{3}(3+cos^3(3x))^{-4/3}\times3cos^2(3x)\times-sin(3x)\times3$
$=3sin(3x)cos^2(3x)(3+cos^3(3x))^{-4/3}$
