(i) To find the equation of the curve, integrate \(\frac{dy}{dx} = 3x^{\frac{1}{2}} - 6\).

\(y = \int (3x^{\frac{1}{2}} - 6) \, dx\)

\(y = \int 3x^{\frac{1}{2}} \, dx - \int 6 \, dx\)

\(y = 2x^{\frac{3}{2}} - 6x + c\)

Using the point (9, 2) to find \(c\):

\(2 = 2(9)^{\frac{3}{2}} - 6(9) + c\)

\(2 = 54 - 54 + c\)

\(c = 2\)

Thus, the equation of the curve is \(y = 2x^{\frac{3}{2}} - 6x + 2\).

(ii) To find the stationary point, set \(\frac{dy}{dx} = 0\):

\(3x^{\frac{1}{2}} - 6 = 0\)

\(3x^{\frac{1}{2}} = 6\)

\(x^{\frac{1}{2}} = 2\)

\(x = 4\)

To determine the nature, find \(\frac{d^2y}{dx^2}\):

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

At \(x = 4\), \(\frac{d^2y}{dx^2} = \frac{3}{2} \times \frac{1}{2} = \frac{3}{4}\)

Since \(\frac{d^2y}{dx^2} > 0\), the stationary point is a minimum.