(i) To express \(\frac{1}{x(2x+3)}\) in partial fractions, assume \(\frac{1}{x(2x+3)} = \frac{A}{x} + \frac{B}{2x+3}\).

Multiply through by \(x(2x+3)\) to get: \(1 = A(2x+3) + Bx\).

Equating coefficients, solve for \(A\) and \(B\):

\(A = \frac{1}{3}\) and \(B = -\frac{2}{3}\).

(ii) Separate variables: \(\frac{dy}{y} = \frac{1}{x(2x+3)} dx\).

Integrate both sides: \(\int \frac{dy}{y} = \int \frac{1}{x} dx - \int \frac{1}{2x+3} dx\).

\(\ln y = \frac{1}{3} \ln x - \frac{1}{3} \ln(2x+3) + C\).

Use initial condition \(y = 1\) when \(x = 1\) to find \(C\):

\(\ln 1 = \frac{1}{3} \ln 1 - \frac{1}{3} \ln 5 + C\) gives \(C = \frac{1}{3} \ln 5\).

Thus, \(\ln y = \frac{1}{3} \ln x - \frac{1}{3} \ln(2x+3) + \frac{1}{3} \ln 5\).

When \(x = 9\), \(\ln y = \frac{1}{3} \ln 9 - \frac{1}{3} \ln 21 + \frac{1}{3} \ln 5\).

Calculate \(y\) to 3 significant figures: \(y = 1.29\).