(i) Substitute \(u = x^{\frac{1}{2}}\), then \(f'(x) = 3u + \frac{3}{u} - 10 = 0\).

Multiply through by \(u\): \(3u^2 - 10u + 3 = 0\).

Factorize: \((3u - 1)(u - 3) = 0\).

So, \(u = \frac{1}{3}\) or \(u = 3\).

Since \(u = x^{\frac{1}{2}}\), \(x = \left(\frac{1}{3}\right)^2 = \frac{1}{9}\) or \(x = 3^2 = 9\).

(ii) Differentiate \(f'(x)\) to find \(f''(x) = \frac{3}{2}x^{-\frac{1}{2}} - \frac{3}{2}x^{-\frac{3}{2}}\).

At \(x = \frac{1}{9}\), \(f''(x) = \frac{3}{2}(3) - \frac{3}{2}(27) = -36 < 0\), indicating a maximum.

At \(x = 9\), \(f''(x) = \frac{3}{2} \times \frac{1}{3} - \frac{3}{2} \times \frac{1}{27} = \frac{4}{9} > 0\), indicating a minimum.