(i) To prove the identity \(\sin^2 \theta \cos^2 \theta \equiv \frac{1}{8}(1 - \cos 4\theta)\), we start by using the double angle identities:

\(\sin 2\theta = 2 \sin \theta \cos \theta\) and \(\cos 2\theta = 2 \cos^2 \theta - 1\).

Thus, \(\sin^2 \theta \cos^2 \theta = \left( \frac{1}{2} \sin 2\theta \right)^2 = \frac{1}{4} \sin^2 2\theta\).

Using \(\sin^2 x = \frac{1}{2}(1 - \cos 2x)\), we have:

\(\sin^2 2\theta = \frac{1}{2}(1 - \cos 4\theta)\).

Therefore, \(\sin^2 \theta \cos^2 \theta = \frac{1}{4} \cdot \frac{1}{2}(1 - \cos 4\theta) = \frac{1}{8}(1 - \cos 4\theta)\).

(ii) To find \(\int_{0}^{\frac{1}{3}\pi} \sin^2 \theta \cos^2 \theta \, d\theta\), substitute the identity:

\(\int_{0}^{\frac{1}{3}\pi} \frac{1}{8}(1 - \cos 4\theta) \, d\theta = \frac{1}{8} \left[ \int_{0}^{\frac{1}{3}\pi} 1 \, d\theta - \int_{0}^{\frac{1}{3}\pi} \cos 4\theta \, d\theta \right]\).

The first integral is:

\(\int_{0}^{\frac{1}{3}\pi} 1 \, d\theta = \left[ \theta \right]_{0}^{\frac{1}{3}\pi} = \frac{1}{3}\pi\).

The second integral is:

\(\int_{0}^{\frac{1}{3}\pi} \cos 4\theta \, d\theta = \left[ \frac{1}{4} \sin 4\theta \right]_{0}^{\frac{1}{3}\pi} = \frac{1}{4} \left( \sin \left( \frac{4}{3}\pi \right) - \sin 0 \right)\).

Since \(\sin \left( \frac{4}{3}\pi \right) = -\frac{\sqrt{3}}{2}\), we have:

\(\frac{1}{4} \left( -\frac{\sqrt{3}}{2} \right) = -\frac{\sqrt{3}}{8}\).

Thus, the integral evaluates to:

\(\frac{1}{8} \left( \frac{1}{3}\pi + \frac{\sqrt{3}}{8} \right)\).