(i) To find stationary points, we differentiate \(y = x^3 + ax^2 + bx\) to get \(\frac{dy}{dx} = 3x^2 + 2ax + b\). For no stationary points, the discriminant of this quadratic must be less than zero: \(b^2 - 4ac < 0\). Here, \(a = 2a\) and \(c = b\), so \(b^2 - 4(3)(b) < 0\) simplifies to \(a^2 < 3b\).

(ii) With \(a = -6\) and \(b = 9\), the equation becomes \(y = x^3 - 6x^2 + 9x\). Differentiating gives \(\frac{dy}{dx} = 3x^2 - 12x + 9\). Setting \(\frac{dy}{dx} < 0\) gives \(3x^2 - 12x + 9 < 0\). Solving \(3(x^2 - 4x + 3) < 0\) gives roots at \(x = 1\) and \(x = 3\). Thus, \(y\) is decreasing for \(1 < x < 3\).