To find the value of x for which the function 3x\(^3\) - 9x\(^2\) is minimum, we can take the derivative of the function and set it equal to zero to find the critical points:

\(\frac{d}{dx}(3x^3 - 9x^2) = 9x^2 - 18x = 0\)

Solving for x, we get:

\(9x(x - 2) = 0\)

This gives us two solutions: x = 0 and x = 2. These are the critical points of the function. To determine whether these points are minima, maxima, or neither, we can use the second derivative test. The second derivative of the function is:

\(\frac{d^2}{dx^2}(3x^3 - 9x^2) = 18x - 18\)

Substituting x = 0 into this expression, we get:

18(0) - 18 = -18

Since this value is negative, the critical point x = 0 is a local maximum.

Substituting x = 2 into this expression, we get:

18(2) - 18 = 18

Since this value is positive, the critical point x = 2 is a local minimum.

Therefore, the value of x for which the function 3x\(^3\) - 9x\(^2\) is minimum is 2.