Answer
$$\frac{dy}{dt}=6\tan(\sin^3t)\sec^2(\sin^3t)\sin^2t\cos t$$
Work Step by Step
$$\frac{dy}{dt}=\frac{d}{dt}\tan^2(\sin^3t)$$
Following the Chain Rule: $$\frac{dy}{dt}=2\tan(\sin^3t)\frac{d}{dt}(\tan(\sin^3t))$$
$$\frac{dy}{dt}=2\tan(\sin^3t)\sec^2(\sin^3t)\frac{d}{dt}(\sin^3t)$$
$$\frac{dy}{dt}=2\tan(\sin^3t)\sec^2(\sin^3t)3\sin^2t\frac{d}{dt}(\sin t)$$
$$\frac{dy}{dt}=6\tan(\sin^3t)\sec^2(\sin^3t)\sin^2t\cos t$$