To find the gradient of the curve, we need to differentiate \(y = \frac{e^{2x}}{1 + e^{2x}}\) with respect to \(x\).

Using the quotient rule, \(\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}\), where \(u = e^{2x}\) and \(v = 1 + e^{2x}\).

First, find \(\frac{du}{dx} = 2e^{2x}\) and \(\frac{dv}{dx} = 2e^{2x}\).

Substitute into the quotient rule:

\(\frac{dy}{dx} = \frac{(1 + e^{2x})(2e^{2x}) - e^{2x}(2e^{2x})}{(1 + e^{2x})^2}\)

Simplify the numerator:

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

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

Now, substitute \(x = \ln 3\):

\(e^{2x} = e^{2 \ln 3} = 3^2 = 9\).

Substitute into the derivative:

\(\frac{dy}{dx} = \frac{2 \times 9}{(1 + 9)^2} = \frac{18}{100} = \frac{9}{50}\)

Thus, the gradient at \(x = \ln 3\) is \(\frac{9}{50}\).