(i) Integrate \(\frac{dy}{dx} = -\frac{k}{x^3}\) to get \(y = -k \frac{x^{-2}}{-2} + c = \frac{k}{2x^2} + c\).

Substitute \((1, 18)\):

\(18 = \frac{k}{2} + c\)

Substitute \((4, 3)\):

\(3 = \frac{k}{32} + c\)

Solving these equations gives \(k = 32\) and \(c = 2\).

Thus, \(y = \frac{16}{x^2} + 2\).

(ii) The area of the shaded region is given by:

\(\int_{1}^{1.6} \left( \frac{16}{x^2} + 2 \right) \, dx\)

Calculate:

\(\left[ -\frac{16}{x} + 2x \right]_{1}^{1.6} = \left[ -10 + 3.2 \right] - \left[ -16 + 2 \right] = 7.2\).