Answer
$$ y' =-36x^2\cot (1-x^3)\csc^2(1-x^3)(1-\csc^2(1-x^3))^5.$$
Work Step by Step
Since $ y=(1-\csc^2(1-x^3))^6$, by using the chain rule: $(f(g(x)))^{\prime}=f^{\prime}(g(x)) g^{\prime}(x)$ and recalling that $(\csc x)'=-\csc x\cot x $, the derivative $ y'$ is given by
$$ y'=6(1-\csc^2(1-x^3))^5(1-\csc^2(1-x^3))'\\
=6(1-\csc^2(1-x^3))^5(-2\csc(1-x^3).\\(-\csc (1-x^3)\cot (1-x^3)(-3x^2)))\\ =-36x^2\cot (1-x^3)\csc^2(1-x^3)(1-\csc^2(1-x^3))^5.$$
