Answer
$$ y'=3\tan^2x \sec^2 x+3x^2\sec^2 x^3.$$
Work Step by Step
Recall that $(\tan x)'=\sec^2 x$.
Recall that $(x^n)'=nx^{n-1}$
Since $ y=\tan^3 x+\tan x^3$, by using the chain rule, the derivative $ y'$ is given by
$$ y'=3\tan^2x \sec^2 x+3x^2\sec^2 x^3.$$
