First, find \(\frac{dx}{dt}\):

\(\frac{dx}{dt} = 2 + 2\cos 2t\).

Next, find \(\frac{dy}{dt}\) using the chain rule:

\(y = \ln(1 - \cos 2t)\).

\(\frac{dy}{dt} = \frac{d}{dt}[\ln(1 - \cos 2t)] = \frac{1}{1 - \cos 2t} \cdot \frac{d}{dt}[-\cos 2t]\).

\(\frac{d}{dt}[-\cos 2t] = 2\sin 2t\).

Thus, \(\frac{dy}{dt} = \frac{2\sin 2t}{1 - \cos 2t}\).

Now, use the formula \(\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt}\):

\(\frac{dy}{dx} = \frac{2\sin 2t}{1 - \cos 2t} \div (2 + 2\cos 2t)\).

Simplify:

\(\frac{dy}{dx} = \frac{2\sin 2t}{(1 - \cos 2t)(2 + 2\cos 2t)}\).

Recognize that \(1 - \cos 2t = 2\sin^2 t\) and \(2 + 2\cos 2t = 4\cos^2 t\).

Thus, \(\frac{dy}{dx} = \frac{2\sin 2t}{2\sin^2 t \cdot 4\cos^2 t} = \frac{1}{\sin 2t} = \csc 2t\).