(i) To find \(\frac{dy}{dx}\), use the quotient rule. Let \(u = 12\) and \(v = 3 - 2x\). Then \(\frac{du}{dx} = 0\) and \(\frac{dv}{dx} = -2\).

Using the quotient rule: \(\frac{dy}{dx} = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} = \frac{(3 - 2x) \cdot 0 - 12 \cdot (-2)}{(3 - 2x)^2} = \frac{24}{(3 - 2x)^2}\).

(ii) Given \(\frac{dy}{dt} = 0.4\) and \(\frac{dx}{dt} = 0.15\), use the chain rule: \(\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt} = \frac{0.4}{0.15} = \frac{8}{3}\).

Equate this to the derivative found in part (i): \(\frac{24}{(3 - 2x)^2} = \frac{8}{3}\).

Cross-multiply to solve for \(x\): \(24 \cdot 3 = 8 \cdot (3 - 2x)^2\).

\(72 = 8(3 - 2x)^2\)

\(9 = (3 - 2x)^2\)

Taking the square root: \(3 - 2x = 3\) or \(3 - 2x = -3\).

Solving these gives \(x = 0\) or \(x = 3\).