(i) To find \(f'(x)\), differentiate \(f(x) = \frac{1}{x+1} + \frac{1}{(x+1)^2}\).

The derivative of \(\frac{1}{x+1}\) is \(-(x+1)^{-2}\).

The derivative of \(\frac{1}{(x+1)^2}\) is \(-2(x+1)^{-3}\).

Thus, \(f'(x) = -(x+1)^{-2} - 2(x+1)^{-3}\).

(ii) Since \(f'(x) < 0\) for \(x > -1\), the function \(f\) is decreasing.

(iii) To find the stationary point of \(g(x) = \frac{1}{x+1} + \frac{1}{(x+1)^2}\), set \(\frac{dy}{dx} = 0\).

\(-\frac{1}{(x+1)^2} - \frac{2}{(x+1)^3} = 0\) or \(-\frac{x^2 - 4x - 3}{(x+1)^4} = 0\).

Multiply by \((x+1)^3\) to get \(-(x+1) - 2 = 0\) or \(-x^2 - 4x - 3 = 0\).

Solving \(-x^2 - 4x - 3 = 0\) gives \(x = -3\).

Substitute \(x = -3\) into \(g(x)\) to find \(y = -1/4\).

Thus, the coordinates of the stationary point are \((-3, -1/4)\).