(i) To find stationary points, set the derivative \(\frac{dy}{dx} = 3x^2 + 2px\) to zero:

\(3x^2 + 2px = 0\)

\(x(3x + 2p) = 0\)

Thus, \(x = 0\) or \(x = -\frac{2p}{3}\).

For \(x = 0\), \(y = 0\), so one stationary point is \((0, 0)\).

For \(x = -\frac{2p}{3}\), substitute back into the equation:

\(y = \left(-\frac{2p}{3}\right)^3 + p\left(-\frac{2p}{3}\right)^2\)

\(y = -\frac{8p^3}{27} + \frac{4p^3}{9} = \frac{4p^3}{27}\)

So the other stationary point is \(\left(-\frac{2p}{3}, \frac{4p^3}{27}\right)\).

(ii) To determine the nature, find the second derivative:

\(\frac{d^2y}{dx^2} = 6x + 2p\)

At \((0, 0)\), \(\frac{d^2y}{dx^2} = 2p > 0\), so it is a minimum.

At \(\left(-\frac{2p}{3}, \frac{4p^3}{27}\right)\), \(\frac{d^2y}{dx^2} = -2p < 0\), so it is a maximum.

(iii) For the curve \(y = x^3 + px^2 + px\), the derivative is:

\(\frac{dy}{dx} = 3x^2 + 2px + p\)

Using the discriminant \(b^2 - 4ac\) for \(3x^2 + 2px + p = 0\):

\((2p)^2 - 4 \cdot 3 \cdot p < 0\)

\(4p^2 - 12p < 0\)

\(4p(p - 3) < 0\)

Thus, \(0 < p < 3\).