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

\(x = (\ln t)^2\) implies \(\frac{dx}{dt} = 2 \ln t \cdot \frac{1}{t} = \frac{2 \ln t}{t}\).

\(y = e^{2-t^2}\) implies \(\frac{dy}{dt} = -2t e^{2-t^2}\).

The gradient \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-2t e^{2-t^2}}{\frac{2 \ln t}{t}} = -t^2 e^{2-t^2} \cdot \frac{t}{2 \ln t} = -\frac{t^3 e^{2-t^2}}{2 \ln t}\).

Substitute \(t = e\):

\(\frac{dy}{dx} = -\frac{e^3 e^{2-e^2}}{2 \ln e} = -\frac{e^3 e^{2-e^2}}{2}\).

Simplifying gives \(-e^{4-e^2}\).

(a) To find the gradient \(\frac{dy}{dx}\), we first find \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\).

\(\frac{dx}{d\theta} = \sec^2 \theta\)

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

Using the chain rule, \(\frac{dy}{dx} = \frac{dy}{d\theta} \div \frac{dx}{d\theta} = \frac{-2 \sin \theta \cos \theta}{\sec^2 \theta} = -2 \sin \theta \cos^3 \theta\).

(b) To find the \(x\)-coordinate of \(P\), we need to find where the derivative \(\frac{dy}{dx}\) is maximized.

Using the product rule, \(\frac{d}{d\theta}(-2 \cos^3 \theta \sin \theta) = -2(\cos^4 \theta + 3 \cos^2 \theta \sin^2 \theta)\).

Setting this derivative to zero gives \(3 \tan^2 \theta = 1\), so \(\tan \theta = \pm \frac{1}{\sqrt{3}}\).

Since \(x = \tan \theta\), the \(x\)-coordinate of \(P\) is \(x = -\frac{1}{\sqrt{3}}\).

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

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

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

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

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

Simplify the expression:

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

Using the double angle identities, \(\cos 2\theta = 1 - 2\sin^2 \theta\) and \(\sin 2\theta = 2\sin \theta \cos \theta\), we have:

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

\(\frac{dy}{dx} = \frac{\cos \theta}{\sin \theta} = \cot \theta\)

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

(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 \(y\):

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

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