To find the stationary points, we need to find where the derivative \(\frac{dy}{dx} = 0\).

Using the product rule, the derivative of \(y = e^{-3x} \tan x\) is:

\(\frac{dy}{dx} = e^{-3x} \cdot \frac{d}{dx}(\tan x) + \tan x \cdot \frac{d}{dx}(e^{-3x})\)

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

Setting \(\frac{dy}{dx} = 0\):

\(e^{-3x} \sec^2 x - 3e^{-3x} \tan x = 0\)

\(\sec^2 x = 3 \tan x\)

\(\frac{1}{\cos^2 x} = 3 \frac{\sin x}{\cos x}\)

\(\cos x = 3 \sin x\)

\(\tan x = \frac{1}{3}\)

Solving \(\tan x = \frac{1}{3}\) within the interval \(-\frac{1}{2}\pi < x < \frac{1}{2}\pi\), we find:

\(x = \arctan(\frac{1}{3}) \approx 0.365\)

Considering the periodicity of the tangent function, the next solution in the interval is:

\(x = \arctan(\frac{1}{3}) + \pi \approx 1.206\)

Thus, the x-coordinates of the stationary points are \(0.365\) and \(1.206\).