Answer: (i) \(\frac{3\sqrt{3}}{2}\)

(ii) \(6\)

We have \(x=2\sin 2t\) and \(y=3\cos 2t\).

Differentiate with respect to \(t\): \(\frac{dx}{dt}=4\cos 2t\) and \(\frac{dy}{dt}=-6\sin 2t\).

So

\(\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{-6\sin 2t}{4\cos 2t}=-\frac{3}{2}\tan 2t\).

At \(t=\frac{\pi}{3}\), we have \(2t=\frac{2\pi}{3}\) and \(\tan\left(\frac{2\pi}{3}\right)=-\sqrt{3}\). Hence

\(\frac{dy}{dx}=-\frac{3}{2}(-\sqrt{3})=\frac{3\sqrt{3}}{2}\).

For the second derivative, differentiate \(\frac{dy}{dx}=-\frac{3}{2}\tan 2t\) with respect to \(t\):

\(\frac{d}{dt}\left(\frac{dy}{dx}\right)=-\frac{3}{2}\cdot 2\sec^2 2t=-3\sec^2 2t\).

Also, \(\frac{dt}{dx}=\frac{1}{dx/dt}=\frac{1}{4\cos 2t}=\frac{1}{4}\sec 2t\).

Therefore

\(\frac{d^2y}{dx^2}=\frac{d}{dt}\left(\frac{dy}{dx}\right)\frac{dt}{dx}=-3\sec^2 2t\cdot \frac{1}{4}\sec 2t=-\frac{3}{4}\sec^3 2t\).

When \(t=\frac{\pi}{3}\), \(\cos\left(\frac{2\pi}{3}\right)=-\frac{1}{2}\), so \(\sec\left(\frac{2\pi}{3}\right)=-2\). Thus

\(\frac{d^2y}{dx^2}=-\frac{3}{4}(-2)^3=6\).