(i) To find the equation of the curve, integrate \(\frac{dy}{dx} = x + \frac{4}{x^2}\).

\(y = \int \left( x + \frac{4}{x^2} \right) \, dx = \int x \, dx + \int \frac{4}{x^2} \, dx\)

\(y = \frac{x^2}{2} - \frac{4}{x} + c\)

Using the point \(P(4, 8)\), substitute \(x = 4\) and \(y = 8\):

\(8 = \frac{4^2}{2} - \frac{4}{4} + c\)

\(8 = 8 - 1 + c\)

\(c = 1\)

Thus, the equation of the curve is \(y = \frac{x^2}{2} - \frac{4}{x} + 1\).

(ii) To find when the gradient has a minimum value, find the second derivative:

\(\frac{d^2y}{dx^2} = \frac{d}{dx} \left( x + \frac{4}{x^2} \right) = 1 - \frac{8}{x^3}\)

Set \(\frac{d^2y}{dx^2} = 0\) to find critical points:

\(1 - \frac{8}{x^3} = 0\)

\(\frac{8}{x^3} = 1\)

\(x^3 = 8\)

\(x = 2\)

Substitute \(x = 2\) into \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = 2 + \frac{4}{2^2} = 2 + 1 = 3\)

Check the nature of the critical point using the third derivative or sign change:

\(\frac{d}{dx} \left( 1 - \frac{8}{x^3} \right) = \frac{24}{x^4}\)

Since \(\frac{24}{x^4} > 0\) for \(x > 0\), the gradient has a minimum value at \(x = 2\).

Thus, the minimum gradient is 3 when \(x = 2\).