To find \(f'(x)\), we differentiate \(f(x) = 2 - \frac{3}{4x-p}\).

Using the chain rule, let \(u = 4x - p\), then \(\frac{du}{dx} = 4\).

The derivative of \(\frac{3}{u}\) with respect to \(u\) is \(-\frac{3}{u^2}\).

Thus, \(\frac{d}{dx} \left( \frac{3}{4x-p} \right) = -3 \cdot \frac{1}{(4x-p)^2} \cdot 4 = \frac{-12}{(4x-p)^2}\).

Therefore, \(f'(x) = 0 - \left( \frac{-12}{(4x-p)^2} \right) = \frac{12}{(4x-p)^2}\).

Since \(\frac{12}{(4x-p)^2} > 0\) for \(x > \frac{p}{4}\), \(f\) is an increasing function.