To find the gradient of the curve, we need to differentiate the equation \(y = 3x^3 - 6x^2 + 4x + 2\) with respect to \(x\).

The derivative is \(\frac{dy}{dx} = 9x^2 - 12x + 4\).

We need to show that \(9x^2 - 12x + 4 \geq 0\) for all \(x\).

Consider the expression \(9x^2 - 12x + 4\).

This can be rewritten as \((3x - 2)^2\).

Since \((3x - 2)^2\) is a square, it is always non-negative.

Therefore, \(9x^2 - 12x + 4 \geq 0\) for all \(x\), and the gradient of the curve is never negative.