To find the stationary points, we first find the derivative \(\frac{dy}{dx}\).

\(\frac{dy}{dx} = 8 + [-2] [(2x-1)^{-2}]\).

Set \(\frac{dy}{dx} = 0\):

\(0 = 8 - \frac{2}{(2x-1)^2}\).

Rearrange to find \((2x-1)^2 = \frac{1}{4}\).

Solving gives \(x = \frac{1}{4}\) and \(x = \frac{3}{4}\).

Next, find the second derivative \(\frac{d^2y}{dx^2} = 8(2x-1)^{-3}\).

Evaluate \(\frac{d^2y}{dx^2}\) at each stationary point:

When \(x = \frac{1}{4}\), \(\frac{d^2y}{dx^2} = -64\), which is less than 0, indicating a maximum.

When \(x = \frac{3}{4}\), \(\frac{d^2y}{dx^2} = 64\), which is greater than 0, indicating a minimum.