Given : \(y = x^{3} - x^{2}\)

\(\frac{\mathrm d y}{\mathrm d x} = 3x^{2} - 2x\)

\(\therefore  \text{The gradient of the tangent at point (x = 2)} = 3(2^{2}) - 2(2) \)

= \(12 - 4 = 8\)

Recall, the tangent and the normal are perpendicular to each other and the product of the gradients of perpendicular lines = -1.

\(\implies \text{the gradient of the normal} = \frac{-1}{8}\)