Answer
$$\frac{dy}{dt}=-\frac{5}{3}\sin\Big(5\sin\Big(\frac{t}{3}\Big)\Big)\cos\Big(\frac{t}{3}\Big)$$
Work Step by Step
$$\frac{dy}{dt}=\frac{d}{dt}\Big(\cos\Big(5\sin\Big(\frac{t}{3}\Big)\Big)\Big)$$
Following the Chain Rule: $$\frac{dy}{dt}=-\sin\Big(5\sin\Big(\frac{t}{3}\Big)\Big)\frac{d}{dt}\Big(5\sin\Big(\frac{t}{3}\Big)$$
$$\frac{dy}{dt}=-\sin\Big(5\sin\Big(\frac{t}{3}\Big)\Big)\times5\cos\Big(\frac{t}{3}\Big)\frac{d}{dt}\Big(\frac{t}{3}\Big)$$
$$\frac{dy}{dt}=-5\sin\Big(5\sin\Big(\frac{t}{3}\Big)\Big)\cos\Big(\frac{t}{3}\Big)\Big(\frac{1}{3}\Big)$$
$$\frac{dy}{dt}=-\frac{5}{3}\sin\Big(5\sin\Big(\frac{t}{3}\Big)\Big)\cos\Big(\frac{t}{3}\Big)$$