Answer:

Differentiate both parametric equations with respect to \(t\):

\(\frac{dx}{dt}=2\cdot 3\cos^{2}t(-\sin t)=-6\cos^{2}t\sin t,\qquad \frac{dy}{dt}=2\cdot 3\sin^{2}t\cos t=6\sin^{2}t\cos t.\)

Hence

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

Now differentiate \(\frac{dy}{dx}\) with respect to \(t\), then divide by \(dx/dt\):

\(\frac{d^{2}y}{dx^{2}}=\frac{\frac{d}{dt}(-\tan t)}{dx/dt}=\frac{-\sec^{2}t}{-6\cos^{2}t\sin t}=\frac{\sec^{2}t}{6\cos^{2}t\sin t}.\)

Since \(1/\cos^{2}t=\sec^{2}t\) and \(1/\sin t=\operatorname{cosec}t\),

\(\frac{d^{2}y}{dx^{2}}=\frac{1}{6}\sec^{4}t\,\operatorname{cosec}t.\)