(i) Differentiate \(x\) and \(y\) with respect to \(\theta\):

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

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

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

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

(ii) To find where the tangent is parallel to the \(y\)-axis, set \(\frac{dx}{d\theta} = 0\):

\(2 \cos \theta + 2 \cos 2\theta = 0\)

\(\cos \theta + \cos 2\theta = 0\)

Using the double angle formula, \(\cos 2\theta = 2\cos^2 \theta - 1\), we get:

\(\cos \theta + 2\cos^2 \theta - 1 = 0\)

\(2\cos^2 \theta + \cos \theta - 1 = 0\)

Solving this quadratic equation for \(\cos \theta\), we find:

\(\cos \theta = \frac{1}{2}\)

Substitute back to find \(x\) and \(y\):

\(x = 2 \sin \theta + \sin 2\theta = \frac{3\sqrt{3}}{2}\)

\(y = 2 \cos \theta + \cos 2\theta = \frac{1}{2}\)