First, find \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\):

\(\frac{dx}{d\theta} = 2\sin 2\theta\)

\(\frac{dy}{d\theta} = 2 + 2\cos 2\theta\)

Now, use the chain rule to find \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{2 + 2\cos 2\theta}{2\sin 2\theta}\)

Simplify the expression:

\(\frac{dy}{dx} = \frac{1 + \cos 2\theta}{\sin 2\theta}\)

Using the double angle identities, \(\cos 2\theta = 1 - 2\sin^2 \theta\) and \(\sin 2\theta = 2\sin \theta \cos \theta\), we have:

\(\frac{dy}{dx} = \frac{1 + (1 - 2\sin^2 \theta)}{2\sin \theta \cos \theta} = \frac{2\cos^2 \theta}{2\sin \theta \cos \theta}\)

\(\frac{dy}{dx} = \frac{\cos \theta}{\sin \theta} = \cot \theta\)