(i) To find the stationary point, we need to find the derivative \(y'\) and set it to zero. Using the quotient rule for \(y = \frac{e^{2x}}{x^3}\), we have:

\(y' = \frac{(2e^{2x})(x^3) - (e^{2x})(3x^2)}{(x^3)^2} = \frac{2e^{2x}x^3 - 3e^{2x}x^2}{x^6}\)

Simplifying, \(y' = \frac{e^{2x}(2x^3 - 3x^2)}{x^6} = \frac{e^{2x}x^2(2x - 3)}{x^6} = \frac{e^{2x}(2x - 3)}{x^4}\)

Setting \(y' = 0\), we solve \(2x - 3 = 0\), giving \(x = \frac{3}{2}\).

(ii) To determine the nature of the stationary point, we test the sign of \(y'\) around \(x = \frac{3}{2}\). For \(x < \frac{3}{2}\), \(2x - 3 < 0\), so \(y' < 0\). For \(x > \frac{3}{2}\), \(2x - 3 > 0\), so \(y' > 0\). This change from negative to positive indicates a minimum point.