To find when the function \(f(x) = (2x - 1)^{\frac{3}{2}} - 6x\) is decreasing, we need to find when its derivative \(f'(x)\) is less than or equal to zero.

First, find the derivative:

\(f'(x) = \left( \frac{3}{2} \right)(2x - 1)^{\frac{1}{2}} \times [2] - [6]\)

Simplifying, we get:

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

Set \(f'(x) \leq 0\):

\(3(2x - 1)^{\frac{1}{2}} - 6 \leq 0\)

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

\((2x - 1)^{\frac{1}{2}} \leq 2\)

Square both sides:

\(2x - 1 \leq 4\)

\(2x \leq 5\)

\(x \leq \frac{5}{2}\)

Thus, the largest value of \(k\) for which \(f\) is a decreasing function is \(\frac{5}{2}\).