(i) Calculate f'(2):

\(f'(2) = 2(2) - \frac{2}{2^2} = 4 - \frac{1}{2} = \frac{7}{2}\)

The gradient of the normal is the negative reciprocal:

\(-\frac{2}{7}\)

Equation of the normal using point-slope form:

\(y - 6 = -\frac{2}{7}(x - 2)\)

(ii) Integrate f'(x) to find f(x):

\(f(x) = \int (2x - \frac{2}{x^2}) \, dx = x^2 + \frac{2}{x} + c\)

Substitute point (2, 6):

\(6 = 4 + 1 + c \Rightarrow c = 1\)

Thus, \(f(x) = x^2 + \frac{2}{x} + 1\)

\((iii) Find stationary points by setting f'(x) = 0:\)

\(2x - \frac{2}{x^2} = 0 \Rightarrow 2x^3 - 2 = 0 \Rightarrow x = 1\)

Determine nature using second derivative:

\(f''(x) = 2 + \frac{4}{x^3}\)

\(f''(1) = 6\) (which is > 0), hence minimum.