First, express \(gf(x)\) in terms of \(x\):

\(gf(x) = g(f(x)) = g(3x - 2)\).

Substitute \(3x - 2\) into \(g(x) = 6x - x^2\):

\(gf(x) = 6(3x - 2) - (3x - 2)^2\).

Expand and simplify:

\(gf(x) = 18x - 12 - (9x^2 - 12x + 4)\).

\(gf(x) = 18x - 12 - 9x^2 + 12x - 4\).

\(gf(x) = -9x^2 + 30x - 16\).

To find the maximum value, differentiate \(gf(x)\):

\(\frac{d}{dx} gf(x) = -18x + 30\).

Set the derivative to zero to find critical points:

\(-18x + 30 = 0\).

\(18x = 30\).

\(x = \frac{5}{3}\).

Substitute \(x = \frac{5}{3}\) back into \(gf(x)\) to find the maximum value:

\(gf\left(\frac{5}{3}\right) = -9\left(\frac{5}{3}\right)^2 + 30\left(\frac{5}{3}\right) - 16\).

\(gf\left(\frac{5}{3}\right) = -9\left(\frac{25}{9}\right) + 50 - 16\).

\(gf\left(\frac{5}{3}\right) = -25 + 50 - 16\).

\(gf\left(\frac{5}{3}\right) = 9\).

Thus, the maximum value of \(gf(x)\) is 9.