(i) At \(x = 2\), \(\frac{d^2y}{dx^2} = \frac{24}{2^3} - 4 = 3 - 4 = -1\), which is negative, indicating a maximum point.

(ii) To find \(\frac{dy}{dx}\), integrate \(\frac{d^2y}{dx^2} = \frac{24}{x^3} - 4\):

\(\frac{dy}{dx} = \int \left( \frac{24}{x^3} - 4 \right) dx = \int 24x^{-3} dx - \int 4 dx\)

\(= -12x^{-2} - 4x + A\)

Given \(\frac{dy}{dx} = 0\) at \(x = 2\), solve for \(A\):

\(0 = -12(2)^{-2} - 4(2) + A\)

\(0 = -3 - 8 + A\)

\(A = 11\)

Thus, \(\frac{dy}{dx} = -12x^{-2} - 4x + 11\).

(iii) Integrate \(\frac{dy}{dx} = -12x^{-2} - 4x + 11\) to find \(y\):

\(y = \int (-12x^{-2} - 4x + 11) dx\)

\(= 12x^{-1} - 2x^2 + 11x + c\)

Given \(y = 13\) when \(x = 1\):

\(13 = 12(1)^{-1} - 2(1)^2 + 11(1) + c\)

\(13 = 12 - 2 + 11 + c\)

\(c = -8\)

Thus, \(y = 12x^{-1} - 2x^2 + 11x - 8\).

At \(x = 2\), \(y = 12(2)^{-1} - 2(2)^2 + 11(2) - 8\)

\(y = 6 - 8 + 22 - 8 = 12\)

Therefore, the coordinates of the stationary point \(P\) are \((2, 12)\).