(a) Integrate \(\frac{d^2y}{dx^2} = -\frac{24}{x^3}\) to find \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = \int -\frac{24}{x^3} \, dx = \frac{12}{x^2} + c\).

Using the stationary point \((-2, 19)\), \(\frac{dy}{dx} = 0\) when \(x = -2\):

\(0 = \frac{12}{(-2)^2} + c \Rightarrow c = -3\).

Thus, \(\frac{dy}{dx} = \frac{12}{x^2} - 3\).

(b) Set \(\frac{dy}{dx} = 0\):

\(\frac{12}{x^2} - 3 = 0 \Rightarrow \frac{12}{x^2} = 3 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2\).

Since \(x = -2\) is already given, the other point is \(x = 2\).

Determine the nature using \(\frac{d^2y}{dx^2} = -\frac{24}{x^3}\):

At \(x = 2\), \(\frac{d^2y}{dx^2} = -\frac{24}{8} = -3\), which is negative, indicating a maximum.

(c) Integrate \(\frac{dy}{dx} = \frac{12}{x^2} - 3\) to find \(y\):

\(y = \int \left( \frac{12}{x^2} - 3 \right) \, dx = -\frac{12}{x} - 3x + d\).

Using \((-2, 19)\):

\(19 = -\frac{12}{-2} - 3(-2) + d \Rightarrow 19 = 6 + 6 + d \Rightarrow d = 7\).

Thus, \(y = \frac{12}{x} - 3x + 7\).

(d) Find \(x\) where \(\frac{dy}{dx} = -\frac{9}{4}\):

\(\frac{12}{x^2} - 3 = -\frac{9}{4} \Rightarrow \frac{12}{x^2} = \frac{3}{4} \Rightarrow x^2 = 16 \Rightarrow x = 4\) (since \(x\) is positive).

Find \(y\) at \(x = 4\):

\(y = \frac{12}{4} - 3(4) + 7 = 3 - 12 + 7 = -2\).

The gradient of the normal is \(\frac{4}{9}\).

Equation of the normal: \(y + 2 = \frac{4}{9}(x - 4)\).

Rearrange to \(4x - 9y - 88 = 0\).