(i) To find \(\frac{dy}{dx}\), we first find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\).

\(\frac{dx}{dt} = \frac{d}{dt}(\ln(\tan t)) = \frac{1}{\tan t} \cdot \sec^2 t = \frac{\sec^2 t}{\tan t}\).

\(\frac{dy}{dt} = \frac{d}{dt}(\sin^2 t) = 2 \sin t \cos t\).

Then, \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2 \sin t \cos t}{\sec^2 t / \tan t} = 2 \sin^2 t \cos^2 t\).

(ii) To find the equation of the tangent at \(x = 0\), we first find \(t\) when \(x = 0\).

\(x = \ln(\tan t) = 0 \Rightarrow \tan t = 1 \Rightarrow t = \frac{\pi}{4}\).

At \(t = \frac{\pi}{4}\), \(y = \sin^2(\frac{\pi}{4}) = \left(\frac{\sqrt{2}}{2}\right)^2 = \frac{1}{2}\).

The gradient \(\frac{dy}{dx}\) at \(t = \frac{\pi}{4}\) is \(\frac{1}{2}\).

The equation of the tangent is \(y - \frac{1}{2} = \frac{1}{2}(x - 0)\).

Simplifying, \(y = \frac{1}{2}x + \frac{1}{2}\).