(i) Differentiate \(y = (kx - 3)^{-1} + (kx - 3)\) to find \(\frac{dy}{dx} = -k(kx - 3)^{-2} + k\). Set \(\frac{dy}{dx} = 0\) to find stationary points:

\(-k(kx - 3)^{-2} + k = 0\)

\((kx - 3)^{2} = 1\)

\(k^{2}x^{2} - 6kx + 8k = 0\)

Solving gives \(x = \frac{2}{k}\) or \(x = \frac{4}{k}\).

To determine the nature, find \(\frac{d^{2}y}{dx^{2}} = 2k^{2}(kx - 3)^{-3}\).

When \(x = \frac{2}{k}\), \(\frac{d^{2}y}{dx^{2}} = (-2k^{2}) < 0\), so it is a maximum.

When \(x = \frac{4}{k}\), \(\frac{d^{2}y}{dx^{2}} = (2k^{2}) > 0\), so it is a minimum.

(ii) For \(k = 1\), the curve is \(y = (x - 3)^{-1} + (x - 3)\). The volume \(V\) is given by:

\(V = \pi \int_{0}^{2} [(x - 3)^{-1} + (x - 3)]^{2} \, dx\)

\(= \pi \int_{0}^{2} [(x - 3)^{-2} + (x - 3)^{2} + 2] \, dx\)

\(= \pi \left[ -(x - 3)^{-1} + \frac{(x - 3)^{3}}{3} + 2x \right]_{0}^{2}\)

\(= \pi \left[ -\frac{1}{3} + 4 - \left(-\frac{1}{3} - 9 + 0\right) \right]\)

\(= \frac{40\pi}{3}\) or 41.9