(i) Integrate \(\frac{d^2y}{dx^2} = 2x - 5\) to find \(\frac{dy}{dx} = x^2 - 5x + c\).

Since the curve has a stationary point at \(x = 3\), \(\frac{dy}{dx} = 0\) when \(x = 3\).

Substitute \(x = 3\) into \(x^2 - 5x + c = 0\):

\(9 - 15 + c = 0\)

\(c = 6\)

Integrate again to find \(y = \frac{x^3}{3} - \frac{5x^2}{2} + 6x + d\).

Use the point (3, 6) to find \(d\):

\(6 = \frac{27}{3} - \frac{45}{2} + 18 + d\)

\(6 = 9 - 22.5 + 18 + d\)

\(d = \frac{1}{2}\)

Thus, the equation of the curve is \(y = \frac{x^3}{3} - \frac{5x^2}{2} + 6x + \frac{1}{2}\).

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

\(x^2 - 5x + 6 = 0\)

Factorize to get \((x - 2)(x - 3) = 0\).

The other stationary point is at \(x = 2\).

(iii) Determine the nature of the stationary points using \(\frac{d^2y}{dx^2} = 2x - 5\):

At \(x = 3\), \(\frac{d^2y}{dx^2} = 2(3) - 5 = 1\) (positive, so minimum).

At \(x = 2\), \(\frac{d^2y}{dx^2} = 2(2) - 5 = -1\) (negative, so maximum).