(ii) Start with \(y = \frac{6}{2x+3}\). To find \(f^{-1}(x)\), interchange \(x\) and \(y\):

\(x = \frac{6}{2y+3}\)

Solve for \(y\):

\(x(2y+3) = 6\)

\(2xy + 3x = 6\)

\(2xy = 6 - 3x\)

\(y = \frac{6 - 3x}{2x}\)

\(f^{-1}(x) = \frac{1}{2}(\frac{6}{x} - 3)\)

The domain of \(f^{-1}\) is \(0 < x \leq 2\) because \(f(x)\) is defined for \(x \geq 0\) and \(f(x) \to 0\) as \(x \to \infty\).

(iii) The graph of \(y = f^{-1}(x)\) is a reflection of \(y = f(x)\) across the line \(y = x\).

(iv) Given \(fg(x) = \frac{3}{2}\), substitute \(g(x) = \frac{1}{2}x\):

\(f\left(\frac{1}{2}x\right) = \frac{3}{2}\)

\(\frac{6}{2\left(\frac{1}{2}x\right) + 3} = \frac{3}{2}\)

\(\frac{6}{x + 3} = \frac{3}{2}\)

Cross-multiply:

\(12 = 3(x + 3)\)

\(12 = 3x + 9\)

\(3 = 3x\)

\(x = 1\)