(a) Start with \(y = \frac{2x}{3x-1}\). Rearrange to find \(x\) in terms of \(y\):

\(y(3x-1) = 2x\)

\(3xy - y = 2x\)

\(3xy - 2x = y\)

\(x(3y - 2) = y\)

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

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

(b) Start with \(\frac{2}{3} + \frac{2}{3(3x-1)}\):

\(\frac{2}{3} + \frac{2}{3(3x-1)} = \frac{2(3x-1) + 2}{3(3x-1)}\)

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

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

(c) The range of \(f\) is \(f(x) > \frac{2}{3}\).