(i) At a stationary point, \(\frac{dy}{dx} = 0\). Given \(\frac{dy}{dx} = ax^2 + a^2 x\), substitute \(x = 3\):

\(0 = 9a + 3a^2\)

Solving gives \(a = -3\).

(ii) Integrate \(\frac{dy}{dx} = -3x^2 + 9x\) to find \(y\):

\(y = \int (-3x^2 + 9x) \, dx = -x^3 + \frac{9x^2}{2} + c\)

Substitute \((3, 9\frac{1}{2})\) to find \(c\):

\(9\frac{1}{2} = -27 + \frac{40}{2} + c\)

\(c = -4\)

Thus, \(y = -x^3 + \frac{9x^2}{2} - 4\).

(iii) Find \(\frac{d^2y}{dx^2}\) to determine the nature of the stationary point:

\(\frac{d^2y}{dx^2} = -6x + 9\)

At \(x = 3\), \(\frac{d^2y}{dx^2} = -9\), which is less than 0, indicating a maximum.