(i) Differentiate \(y = \frac{12}{x^2 + 3}\) using the chain rule. Let \(u = x^2 + 3\), then \(y = \frac{12}{u}\). The derivative \(\frac{dy}{du} = -\frac{12}{u^2}\). Since \(u = x^2 + 3\), \(\frac{du}{dx} = 2x\). Therefore, \(\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} = -\frac{12}{(x^2 + 3)^2} \times 2x\).

(ii) At \(x = 1\), the slope \(m = \frac{dy}{dx} = -\frac{3}{2}\). The slope of the normal is the negative reciprocal, \(\frac{2}{3}\). The equation of the normal is \(y - 3 = \frac{2}{3}(x - 1)\).

(iii) Use the chain rule: \(\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}\). Given \(\frac{dx}{dt} = 0.012\) and \(\frac{dy}{dx} = -\frac{3}{2}\) at \(x = 1\), \(\frac{dy}{dt} = -\frac{3}{2} \times 0.012 = -0.018\).