(i) To find the stationary points, set \(\frac{dy}{dx} = 0\). First, find \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = x - 6x^{\frac{1}{2}} + 8\).

Set \(\frac{dy}{dx} = 0\):

\(x - 6x^{\frac{1}{2}} + 8 = 0\).

Let \(u = x^{\frac{1}{2}}\), then \(u^2 = x\). The equation becomes:

\(u^2 - 6u + 8 = 0\).

Solving the quadratic equation gives \(u = 4\) or \(u = 2\).

Thus, \(x = 16\) or \(x = 4\).

(ii) Differentiate \(\frac{dy}{dx} = x - 6x^{\frac{1}{2}} + 8\) to find \(\frac{d^2y}{dx^2}\):

\(\frac{d^2y}{dx^2} = 1 - 3x^{-\frac{1}{2}}\).

(iii) Evaluate \(\frac{d^2y}{dx^2}\) at the stationary points:

When \(x = 16\), \(\frac{d^2y}{dx^2} = 1 - 3 \times 16^{-\frac{1}{2}} = \frac{1}{4} > 0\), indicating a minimum point.

When \(x = 4\), \(\frac{d^2y}{dx^2} = 1 - 3 \times 4^{-\frac{1}{2}} = -\frac{1}{2} < 0\), indicating a maximum point.