(i) Differentiate \(y = 2 + \frac{3}{2x - 1}\) using the chain rule:

Let \(u = 2x - 1\), then \(\frac{du}{dx} = 2\).

\(y = 2 + \frac{3}{u}\), so \(\frac{dy}{du} = -\frac{3}{u^2}\).

Thus, \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = -\frac{3}{(2x-1)^2} \times 2\).

(ii) The curve has no stationary points because \(\frac{dy}{dx} = 0\) is impossible as the derivative is never zero.

(iii) At \(x = 2\), \(\frac{dy}{dx} = \frac{-6}{9} = -\frac{2}{3}\) and \(y = 3\).

The slope of the normal is the negative reciprocal: \(m = \frac{9}{6} = \frac{3}{2}\).

The equation of the normal is \(y - 3 = \frac{3}{2}(x - 2)\).

Substitute \(x = 0\) to find \(y = 0\), showing it passes through the origin.

(iv) Given \(\frac{dx}{dt} = -0.06\), find \(\frac{dy}{dt}\):

\(\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt} = -\frac{2}{3} \times -0.06 = 0.04\).