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

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

\(\Rightarrow 3(3x-k)^2 = k\)

\(\Rightarrow 3x-k = \pm \sqrt{k}\)

\(\Rightarrow x = \frac{k \pm \sqrt{k}}{3}\)

(b) Substitute \(x = a\) when \(k = 4\) into \(f(x) = 4(3x-4)^{-1} + 3x\):

\(a = \frac{4 \pm \sqrt{4}}{3} \Rightarrow a = 2\)

Find the second derivative \(f''(x) = 72(3x-4)^{-3}\), which is positive at \(x = 2\), indicating a minimum.

(c) Substitute \(k = -1\) into \(g(x) = -(3x+1)^{-1} + 3x\) to find \(g'(x) = \frac{3}{(3x+1)^2} + 3\), which is always positive, indicating \(g\) is an increasing function.