Answer
$$ y' =168x^3(5-2x^4)^6(9-(5-2x^4)^7)^2.$$
Work Step by Step
Since $ y=(9-(5-2x^4)^7)^3$, then by using the chain rule: $(f(g(x)))^{\prime}=f^{\prime}(g(x)) g^{\prime}(x)$, the derivative $ y'$ is given by
$$ y'=3(9-(5-2x^4)^7)^2 (9-(5-2x^4)^7)'
\\=3(9-(5-2x^4)^7)^2(-7(5-2x^4)^6(-8x^3))\\=168x^3(5-2x^4)^6(9-(5-2x^4)^7)^2.$$
