(i) To determine whether the stationary point is a maximum or minimum, evaluate the second derivative at \(x = 3\):

\(f''(3) = 36 \times 3^{-3} = \frac{36}{27} = \frac{4}{3}\).

Since \(f''(3) > 0\), the stationary point at \((3, 7)\) is a minimum.

(ii) To find \(f'(x)\), integrate \(f''(x)\):

\(f''(x) = 36x^{-3}\) implies \(f'(x) = \int 36x^{-3} \, dx = -18x^{-2} + c\).

Given that \(f'(3) = 0\), substitute to find \(c\):

\(0 = -18 \times 3^{-2} + c\)

\(0 = -2 + c\)

\(c = 2\)

Thus, \(f'(x) = -18x^{-2} + 2\).

To find \(f(x)\), integrate \(f'(x)\):

\(f(x) = \int (-18x^{-2} + 2) \, dx = 18x^{-1} + 2x + k\).

Given that \(f(3) = 7\), substitute to find \(k\):

\(7 = 18 \times 3^{-1} + 2 \times 3 + k\)

\(7 = 6 + 6 + k\)

\(k = -5\)

Thus, \(f(x) = 18x^{-1} + 2x - 5\).