To find the maximum value, we can use the second derivative test where, given \(f(x)\), the second derivative < 0, makes it a maximum value.

\(x(x + 1)^{2} = x(x^{2} + 2x + 1) = x^{3} + 2x^{2} + x\)

\(\frac{\mathrm d y}{\mathrm d x} = 3x^{2} + 4x + 1 = 0\)

Solving, we have \( x = \frac{-1}{3}\) or \(-1\).

\(\frac{\mathrm d^{2} y}{\mathrm d x^{2}} = 6x + 4\)

When \(x = \frac{-1}{3}, \frac{\mathrm d^{2} y}{\mathrm d x^{2}} = 2 > 0\)

When \(x = -1, \frac{\mathrm d^{2} y}{\mathrm d x^{2}} = -2 < 0\)

At maximum value of x being -1, \(y = -1(-1 + 1)^{2} = 0\)