Answer
$$ y'= 6(6t-60t^{-4}) (3t^2+20t^{-3})^5 .$$
Work Step by Step
Recall that $(x^n)'=nx^{n-1}$
Since $ y=(3t^2+20t^{-3})^6$, then by the chain rule the derivative $ y'$ is given by
$$ y'=6 (3t^2+20t^{-3})^5(6t-60t^{-4})=6(6t-60t^{-4}) (3t^2+20t^{-3})^5 .$$
