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

Using the quotient rule for differentiation, \(\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}\), where \(u = e^{3x}\) and \(v = \tan \frac{1}{2}x\).

First, find \(\frac{du}{dx} = 3e^{3x}\) and \(\frac{dv}{dx} = \frac{1}{2} \sec^2 \frac{1}{2}x\).

Substitute into the quotient rule:

\(\frac{dy}{dx} = \frac{\tan \frac{1}{2}x \cdot 3e^{3x} - e^{3x} \cdot \frac{1}{2} \sec^2 \frac{1}{2}x}{\tan^2 \frac{1}{2}x}\).

Simplify and set \(\frac{dy}{dx} = 0\):

\(3\tan \frac{1}{2}x = \frac{1}{2} \sec^2 \frac{1}{2}x\).

Rearrange to form a quadratic in \(\tan \frac{1}{2}x\):

\(6\tan^2 \frac{1}{2}x = \sec^2 \frac{1}{2}x\).

Using \(\sec^2 \theta = 1 + \tan^2 \theta\), substitute:

\(6\tan^2 \frac{1}{2}x = 1 + \tan^2 \frac{1}{2}x\).

Solve for \(\tan \frac{1}{2}x\):

\(5\tan^2 \frac{1}{2}x = 1\).

\(\tan \frac{1}{2}x = \pm \frac{1}{\sqrt{5}}\).

Find \(x\) in the interval \(0 < x < \pi\):

\(x = 2 \arctan \left( \frac{1}{\sqrt{5}} \right) \approx 0.340\).

\(x = 2 \arctan \left( -\frac{1}{\sqrt{5}} \right) + \pi \approx 2.802\).

Thus, the \(x\)-coordinates of the stationary points are \(0.340\) and \(2.802\).