Let \(y = \frac{15}{2x+3}\).

Rearrange to solve for \(x\):

\(2x + 3 = \frac{15}{y}\)

\(2x = \frac{15}{y} - 3\)

\(x = \frac{15 - 3y}{2y}\)

Thus, \(f^{-1}(x) = \frac{15 - 3x}{2x}\).

For the range of \(f^{-1}\), since \(0 \leq x \leq 6\) for \(f(x)\), the range of \(f^{-1}\) is \(0 \leq f^{-1}(x) \leq 6\).

For the domain of \(f^{-1}\), solve \(0 \leq \frac{15}{2x+3} \leq 6\) for \(x\):

\(1 \leq x \leq 5\).