(i) To find the \(x\)-coordinate of \(A\), set \(\frac{dy}{dx} = 0\):

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

\(2x = \frac{2}{x^3}\)

\(x^4 = 1\)

Thus, \(x = 1\) at \(A\).

(ii) To find \(f(x)\), integrate \(\frac{dy}{dx} = 2x - \frac{2}{x^3}\):

\(f(x) = \int (2x - \frac{2}{x^3}) \, dx\)

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

Using the point \(\left(4, \frac{189}{16}\right)\):

\(\frac{189}{16} = 16 + \frac{1}{16} + c\)

\(c = -\frac{17}{4}\)

Thus, \(f(x) = x^2 + \frac{1}{x^2} - \frac{17}{4}\).

(iii) To find the \(x\)-coordinates of \(B\) and \(C\), solve \(f(x) = 0\):

\(x^2 + \frac{1}{x^2} - \frac{17}{4} = 0\)

Multiply by \(4x^2\):

\(4x^4 - 17x^2 + 4 = 0\)

\((4x^2 - 1)(x^2 - 4) = 0\)

\(x = \frac{1}{2}, 2\)

(iv) To find the area of the shaded region, integrate \(f(x)\) from \(x = \frac{1}{2}\) to \(x = 2\):

\(\int_{1/2}^{2} \left(x^2 + \frac{1}{x^2} - \frac{17}{4}\right) \, dx\)

\(= \left[ \frac{x^3}{3} - \frac{1}{x} - \frac{17x}{4} \right]_{1/2}^{2}\)

\(= \left( \frac{8}{3} - \frac{1}{2} - \frac{17}{2} \right) - \left( \frac{1}{24} - 2 - \frac{17}{8} \right)\)

Area = \(\frac{9}{4}\)