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

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

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

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

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

Simplify the expression:

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

Using the identity \(\sin 2\theta = 2\sin \theta \cos \theta\) and \(1 + \cos 2\theta = 2\cos^2 \theta\), we have:

\(\frac{dy}{dx} = \frac{2\sin \theta \cos \theta}{2\cos^2 \theta} = \frac{\sin \theta}{\cos \theta} = \tan \theta\)