(i) To find \(\frac{dy}{dx}\), use the product rule: \(\frac{d}{dx}[u v] = u'v + uv'\).

Let \(u = \sin(x + \frac{1}{3}\pi)\) and \(v = \cos x\).

Then \(u' = \cos(x + \frac{1}{3}\pi)\) and \(v' = -\sin x\).

Thus, \(\frac{dy}{dx} = \cos(x + \frac{1}{3}\pi) \cos x - \sin(x + \frac{1}{3}\pi) \sin x\).

(ii) At stationary points, \(\frac{dy}{dx} = 0\).

Using the identity \(\cos(A + B) = \cos A \cos B - \sin A \sin B\), we have:

\(\cos(x + \frac{1}{3}\pi) \cos x - \sin(x + \frac{1}{3}\pi) \sin x = \cos(2x + \frac{1}{3}\pi)\).

Therefore, \(\cos(2x + \frac{1}{3}\pi) = 0\).

(iii) Solve \(\cos(2x + \frac{1}{3}\pi) = 0\).

This implies \(2x + \frac{1}{3}\pi = \frac{\pi}{2} + n\pi\), where \(n\) is an integer.

For \(n = 0\), \(2x + \frac{1}{3}\pi = \frac{\pi}{2}\).

\(2x = \frac{\pi}{2} - \frac{1}{3}\pi = \frac{1}{6}\pi\).

\(x = \frac{1}{12}\pi\).

For \(n = 1\), \(2x + \frac{1}{3}\pi = \frac{3\pi}{2}\).

\(2x = \frac{3\pi}{2} - \frac{1}{3}\pi = \frac{7}{6}\pi\).

\(x = \frac{7}{12}\pi\).

Thus, the exact \(x\)-coordinates of the stationary points are \(x = \frac{1}{12}\pi\) and \(x = \frac{7}{12}\pi\).