(i) Differentiate \(x = \ln \cos \theta\) with respect to \(\theta\):

\(\frac{dx}{d\theta} = -\frac{\sin \theta}{\cos \theta} = -\tan \theta\).

Differentiate \(y = 3\theta - \tan \theta\) with respect to \(\theta\):

\(\frac{dy}{d\theta} = 3 - \sec^2 \theta\).

Use the chain rule to find \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = \frac{dy}{d\theta} \div \frac{dx}{d\theta} = \frac{3 - \sec^2 \theta}{-\tan \theta}\).

Simplify to obtain:

\(\frac{dy}{dx} = \frac{\tan^2 \theta - 2}{\tan \theta}\).

(ii) The gradient of the normal is \(-1\), so the gradient of the tangent is \(1\).

Set \(\frac{dy}{dx} = 1\):

\(\frac{\tan^2 \theta - 2}{\tan \theta} = 1\).

Solve the equation:

\(\tan^2 \theta - \tan \theta - 2 = 0\).

Factorize to find \(\tan \theta = 1\) or \(\tan \theta = -2\).

For \(\tan \theta = 1\), \(\theta = \frac{\pi}{4}\).

Substitute \(\theta = \frac{\pi}{4}\) into \(y = 3\theta - \tan \theta\):

\(y = 3 \times \frac{\pi}{4} - 1 = \frac{3\pi}{4} - 1\).

Thus, \(y = -1\).