To find the gradient of the curve, we need to differentiate \(y = \frac{12}{x^2 - 4x}\) with respect to \(x\).

Using the quotient rule, \(\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}\), where \(u = 12\) and \(v = x^2 - 4x\).

First, find \(\frac{du}{dx} = 0\) and \(\frac{dv}{dx} = 2x - 4\).

Applying the quotient rule:

\(\frac{dy}{dx} = \frac{(x^2 - 4x)(0) - 12(2x - 4)}{(x^2 - 4x)^2}\)

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

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

Now, substitute \(x = 3\) into the derivative:

\(\frac{dy}{dx} = -12(3^2 - 4 \times 3)^{-2} \times (2 \times 3 - 4)\)

\(= -12(9 - 12)^{-2} \times (6 - 4)\)

\(= -12(-3)^{-2} \times 2\)

\(= -12 \times \frac{1}{9} \times 2\)

\(= -\frac{24}{9}\)

\(= -\frac{8}{3}\)