(i) Differentiate \(y = \frac{8}{x} + 2x\) with respect to \(x\):

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

Differentiate again to find \(\frac{d^2y}{dx^2}\):

\(\frac{d^2y}{dx^2} = \frac{d}{dx}(-8x^{-2}) = 16x^{-3} = \frac{16}{x^3}\).

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

\(-\frac{8}{x^2} + 2 = 0 \Rightarrow 2x^2 - 8 = 0 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2\).

For \(x = 2\), \(y = \frac{8}{2} + 2(2) = 4 + 4 = 8\).

For \(x = -2\), \(y = \frac{8}{-2} + 2(-2) = -4 - 4 = -8\).

Check the nature of each stationary point using \(\frac{d^2y}{dx^2}\):

At \(x = 2\), \(\frac{d^2y}{dx^2} = \frac{16}{2^3} = 2 > 0\), hence a minimum.

At \(x = -2\), \(\frac{d^2y}{dx^2} = \frac{16}{(-2)^3} = -2 < 0\), hence a maximum.