(i) The gradient of the curve is given by \(\frac{dy}{dx} = kxy\), where \(k\) is a constant. At \((1, 2)\), \(\frac{dy}{dx} = 4\), so \(4 = k \cdot 1 \cdot 2\), giving \(k = 2\). Thus, \(\frac{dy}{dx} = 2xy\).

Separate variables: \(\frac{1}{y} dy = 2x dx\).

Integrate both sides: \(\int \frac{1}{y} dy = \int 2x dx\).

This gives \(\ln |y| = x^2 + C\).

Exponentiate: \(y = e^{x^2 + C}\).

At \((1, 2)\), \(2 = e^{1^2 + C}\), so \(2 = e^{1 + C}\), giving \(C = \ln 2 - 1\).

Thus, \(y = e^{x^2 + \ln 2 - 1} = 2e^{x^2 - 1}\).

(ii) The gradient at \((-1, 2)\) is \(\frac{dy}{dx} = 2(-1)(2) = -4\).