(i) To find the equation of the curve, integrate \(\frac{dy}{dx} = x^{\frac{1}{2}} - x^{-\frac{1}{2}}\).

\(y = \int (x^{\frac{1}{2}} - x^{-\frac{1}{2}}) \, dx = \frac{2}{3}x^{\frac{3}{2}} - 2x^{\frac{1}{2}} + c\).

Substitute the point \((4, \frac{2}{3})\) into the equation:

\(\frac{2}{3} = \frac{2}{3}(4)^{\frac{3}{2}} - 2(4)^{\frac{1}{2}} + c\).

\(\frac{2}{3} = \frac{16}{3} - 4 + c\).

\(c = -\frac{2}{3}\).

Thus, the equation of the curve is \(y = \frac{2}{3}x^{\frac{3}{2}} - 2x^{\frac{1}{2}} - \frac{2}{3}\).

(ii) Differentiate \(\frac{dy}{dx} = x^{\frac{1}{2}} - x^{-\frac{1}{2}}\) to find \(\frac{d^2y}{dx^2}\).

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

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

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

\(x = 1\).

Substitute \(x = 1\) into the equation of the curve:

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

Thus, the stationary point is \((1, -2)\).

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

\(\frac{d^2y}{dx^2} \bigg|_{x=1} = \frac{1}{2}(1)^{-\frac{1}{2}} + \frac{1}{2}(1)^{-\frac{3}{2}} = 1 > 0\).

Hence, the stationary point is a minimum.