(i) To find \(\frac{dy}{dx}\), differentiate \(y = (2x - 1)^{-1} + 2x\).

Using the chain rule, \(\frac{d}{dx}((2x - 1)^{-1}) = -1(2x - 1)^{-2} \cdot 2 = -2(2x - 1)^{-2}\).

The derivative of \(2x\) is \(2\).

Thus, \(\frac{dy}{dx} = -2(2x - 1)^{-2} + 2\).

For \(\frac{d^2y}{dx^2}\), differentiate \(\frac{dy}{dx} = -2(2x - 1)^{-2} + 2\).

Using the chain rule again, \(\frac{d}{dx}(-2(2x - 1)^{-2}) = -2 \cdot -2(2x - 1)^{-3} \cdot 2 = 8(2x - 1)^{-3}\).

Thus, \(\frac{d^2y}{dx^2} = 8(2x - 1)^{-3}\).

(ii) To find stationary points, set \(\frac{dy}{dx} = 0\).

\(-2(2x - 1)^{-2} + 2 = 0\)

\(-2(2x - 1)^{-2} = -2\)

\((2x - 1)^{-2} = 1\)

\((2x - 1)^2 = 1\)

\(2x - 1 = \pm 1\)

\(2x = 2 \Rightarrow x = 1\)

\(2x = 0 \Rightarrow x = 0\)

For \(x = 0\), \(\frac{d^2y}{dx^2} = 8(2 \cdot 0 - 1)^{-3} = -8\), indicating a maximum.

For \(x = 1\), \(\frac{d^2y}{dx^2} = 8(2 \cdot 1 - 1)^{-3} = 8\), indicating a minimum.