(i) To find the equation of the curve, integrate \(\frac{dy}{dx} = (5x - 1)^{\frac{1}{2}} - 2\).

Integrate \((5x - 1)^{\frac{1}{2}}\) using substitution:

Let \(u = 5x - 1\), then \(du = 5 \, dx\), so \(dx = \frac{du}{5}\).

\(\int (5x - 1)^{\frac{1}{2}} \, dx = \int u^{\frac{1}{2}} \cdot \frac{du}{5} = \frac{1}{5} \cdot \frac{2}{3} u^{\frac{3}{2}} + C = \frac{2}{15} (5x - 1)^{\frac{3}{2}} + C\).

Integrate \(-2\) to get \(-2x + D\).

Thus, \(y = \frac{2}{15} (5x - 1)^{\frac{3}{2}} - 2x + C\).

Using the point (2, 3):

\(3 = \frac{2}{15} (5 \times 2 - 1)^{\frac{3}{2}} - 2 \times 2 + C\).

\(3 = \frac{2}{15} \times 27 - 4 + C\).

\(3 = \frac{54}{15} - 4 + C\).

\(3 = \frac{18}{5} - 4 + C\).

\(C = 7 - \frac{18}{5}\).

\(C = \frac{17}{5}\).

Thus, \(y = \frac{2(5x-1)^{\frac{3}{2}}}{15} - 2x + \frac{17}{5}\).

(ii) Differentiate \(\frac{dy}{dx} = (5x - 1)^{\frac{1}{2}} - 2\) to find \(\frac{d^2y}{dx^2}\).

\(\frac{d^2y}{dx^2} = \frac{d}{dx}((5x - 1)^{\frac{1}{2}})\).

Using the chain rule: \(\frac{1}{2}(5x-1)^{-\frac{1}{2}} \times 5\).

\(\frac{d^2y}{dx^2} = \frac{5}{2}(5x-1)^{-\frac{1}{2}}\).

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

\((5x - 1)^{\frac{1}{2}} - 2 = 0\).

\((5x - 1)^{\frac{1}{2}} = 2\).

\(5x - 1 = 4\).

\(5x = 5\).

\(x = 1\).

Substitute \(x = 1\) into \(y = \frac{2(5x-1)^{\frac{3}{2}}}{15} - 2x + \frac{17}{5}\):

\(y = \frac{2(5 \times 1 - 1)^{\frac{3}{2}}}{15} - 2 \times 1 + \frac{17}{5}\).

\(y = \frac{2 \times 8}{15} - 2 + \frac{17}{5}\).

\(y = \frac{16}{15} - 2 + \frac{17}{5}\).

\(y = \frac{16}{15} - \frac{30}{15} + \frac{51}{15}\).

\(y = \frac{37}{15}\).

Stationary point is \((1, \frac{37}{15})\).

Check the nature using \(\frac{d^2y}{dx^2}\):

\(\frac{d^2y}{dx^2} = \frac{5}{2}(5 \times 1 - 1)^{-\frac{1}{2}} = \frac{5}{2} \times \frac{1}{2} = \frac{5}{4}\).

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