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

\(\frac{dy}{dx} = \frac{d}{dx}(x^2) + \frac{d}{dx}\left(\frac{2}{x}\right)\)

\(\frac{dy}{dx} = 2x - \frac{2}{x^2}\)

To find \(\frac{d^2y}{dx^2}\), differentiate \(\frac{dy}{dx} = 2x - \frac{2}{x^2}\):

\(\frac{d^2y}{dx^2} = \frac{d}{dx}(2x) + \frac{d}{dx}\left(-\frac{2}{x^2}\right)\)

\(\frac{d^2y}{dx^2} = 2 + \frac{4}{x^3}\)

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

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

\(2x = \frac{2}{x^2}\)

\(2x^3 = 2\)

\(x^3 = 1\)

\(x = 1\)

Substitute \(x = 1\) into \(y = x^2 + \frac{2}{x}\):

\(y = 1^2 + \frac{2}{1} = 3\)

Stationary point is \((1, 3)\).

To determine the nature, evaluate \(\frac{d^2y}{dx^2}\) at \(x = 1\):

\(\frac{d^2y}{dx^2} = 2 + \frac{4}{1^3} = 6\)

Since \(\frac{d^2y}{dx^2} > 0\), the stationary point is a minimum.