(a) To find \(f'(x)\), use the quotient rule: \(\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}\).

Let \(u = \cos x\) and \(v = 1 + \sin x\). Then \(u' = -\sin x\) and \(v' = \cos x\).

So, \(f'(x) = \frac{-\sin x (1 + \sin x) - \cos x (\cos x)}{(1 + \sin x)^2}\).

Simplify: \(f'(x) = \frac{-\sin x - \sin^2 x - \cos^2 x}{(1 + \sin x)^2}\).

Using \(\sin^2 x + \cos^2 x = 1\), we have \(f'(x) = \frac{-\sin x - 1}{(1 + \sin x)^2}\).

Since \(-\sin x - 1 < 0\) and \((1 + \sin x)^2 > 0\) for \(-\frac{1}{2}\pi < x < \frac{3}{2}\pi\), \(f'(x) < 0\).

(b) To find \(\int_{\frac{1}{6}\pi}^{\frac{1}{2}\pi} f(x) \, dx\), note that \(f(x) = \frac{\cos x}{1 + \sin x}\).

Use substitution: let \(u = 1 + \sin x\), then \(du = \cos x \, dx\).

The integral becomes \(\int \frac{1}{u} \, du = \ln |u| + C\).

Evaluate from \(x = \frac{1}{6}\pi\) to \(x = \frac{1}{2}\pi\):

\(u(\frac{1}{6}\pi) = 1 + \frac{1}{2} = \frac{3}{2}\) and \(u(\frac{1}{2}\pi) = 1 + 1 = 2\).

Thus, \(\int_{\frac{1}{6}\pi}^{\frac{1}{2}\pi} f(x) \, dx = \ln 2 - \ln \frac{3}{2} = \ln \frac{4}{3}\).