(i) To find the stationary point, we first find the derivative of the function \(y = e^x + 4e^{-2x}\).

The derivative is \(\frac{dy}{dx} = e^x - 8e^{-2x}\).

Set the derivative to zero to find the stationary point: \(e^x - 8e^{-2x} = 0\).

Rearrange to get \(e^x = 8e^{-2x}\).

Multiply both sides by \(e^{2x}\) to clear the fraction: \(e^{3x} = 8\).

Take the natural logarithm of both sides: \(3x = \ln 8\).

Solve for \(x\): \(x = \frac{\ln 8}{3} = \ln 2\).

(ii) To determine the nature of the stationary point, we examine the second derivative.

The second derivative is \(\frac{d^2y}{dx^2} = e^x + 16e^{-2x}\).

Substitute \(x = \ln 2\) into the second derivative: \(\frac{d^2y}{dx^2} = e^{\ln 2} + 16e^{-2\ln 2} = 2 + 16 \times \frac{1}{4} = 2 + 4 = 6\).

Since \(\frac{d^2y}{dx^2} > 0\), the stationary point is a minimum.