(a) To find \(\frac{d}{d\theta}(\cot^2 \theta)\), use the chain rule:

\(\frac{d}{d\theta}(\cot^2 \theta) = 2 \cot \theta \cdot \frac{d}{d\theta}(\cot \theta)\).

Given \(\frac{d}{d\theta}(\cot \theta) = -\csc^2 \theta\), substitute:

\(\frac{d}{d\theta}(\cot^2 \theta) = 2 \cot \theta (-\csc^2 \theta) = -\frac{2 \cot \theta}{\sin^2 \theta}\).

(b) Start with the differential equation:

\(x \sin^2 \theta \frac{dx}{d\theta} = \tan^2 \theta - 2 \cot \theta\).

Separate variables:

\(\int x \, dx = \int \frac{\tan^2 \theta - 2 \cot \theta}{\sin^2 \theta} \, d\theta\).

Integrate both sides:

\(\frac{1}{2}x^2 = \int \frac{\tan^2 \theta}{\sin^2 \theta} \, d\theta - \int \frac{2 \cot \theta}{\sin^2 \theta} \, d\theta\).

\(\frac{1}{2}x^2 = \tan \theta + \cot^2 \theta + C\).

Use the initial condition \(x = 2\) when \(\theta = \frac{1}{4}\pi\):

\(\frac{1}{2}(2)^2 = \tan \frac{1}{4}\pi + \cot^2 \frac{1}{4}\pi + C\).

\(2 = 1 + 1 + C \Rightarrow C = 0\).

Thus, \(\frac{1}{2}x^2 = \tan \theta + \cot^2 \theta\).

Substitute \(\theta = \frac{1}{6}\pi\):

\(\frac{1}{2}x^2 = \tan \frac{1}{6}\pi + \cot^2 \frac{1}{6}\pi\).

Calculate \(x\):

\(x = \sqrt{2(\tan \frac{1}{6}\pi + \cot^2 \frac{1}{6}\pi)} \approx 2.67\).