To find the stationary point, we need to find where the derivative \(\frac{dy}{dx}\) is zero.

Given \(y = \frac{e^{2x}}{4 + e^{3x}}\), use the quotient rule: \(\frac{dy}{dx} = \frac{(4 + e^{3x})(2e^{2x}) - e^{2x}(3e^{3x})}{(4 + e^{3x})^2}\).

Simplify the numerator: \((4 + e^{3x})(2e^{2x}) - e^{2x}(3e^{3x}) = 8e^{2x} + 2e^{5x} - 3e^{5x} = 8e^{2x} - e^{5x}\).

Set the derivative to zero: \(8e^{2x} - e^{5x} = 0\).

Factor out \(e^{2x}\): \(e^{2x}(8 - e^{3x}) = 0\).

Since \(e^{2x} \neq 0\), solve \(8 - e^{3x} = 0\) to get \(e^{3x} = 8\).

Taking the natural logarithm: \(3x = \ln 8\), so \(x = \frac{\ln 8}{3} = \ln 2\).

Substitute \(x = \ln 2\) back into the original equation to find \(y\):

\(y = \frac{e^{2 \ln 2}}{4 + e^{3 \ln 2}} = \frac{4}{4 + 8} = \frac{4}{12} = \frac{1}{3}\).

Thus, the coordinates of the stationary point are \((\ln 2, \frac{1}{3})\).