(i) Differentiate \(y = x^3 + 3x^2 - 9x + k\) with respect to \(x\):

\(\frac{dy}{dx} = 3x^2 + 6x - 9\).

(ii) Stationary points occur where \(\frac{dy}{dx} = 0\):

\(3x^2 + 6x - 9 = 0\).

Factorize the quadratic equation:

\(3(x^2 + 2x - 3) = 0\).

\(3(x + 3)(x - 1) = 0\).

Thus, \(x = -3\) or \(x = 1\).

(iii) For the curve to have a stationary point on the \(x\)-axis, \(y = 0\) at the stationary points:

Substitute \(x = -3\) into \(y = x^3 + 3x^2 - 9x + k\):

\((-3)^3 + 3(-3)^2 - 9(-3) + k = 0\).

\(-27 + 27 + 27 + k = 0\).

\(k = -27\).

Substitute \(x = 1\) into \(y = x^3 + 3x^2 - 9x + k\):

\(1^3 + 3(1)^2 - 9(1) + k = 0\).

\(1 + 3 - 9 + k = 0\).

\(k = 5\).

Thus, the values of \(k\) are \(-27\) and \(5\).