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

Using the product rule, the derivative is:

\(\frac{dy}{dx} = -5e^{-5x} \tan^2 x + 2e^{-5x} \tan x \sec^2 x\).

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

\(-5e^{-5x} \tan^2 x + 2e^{-5x} \tan x \sec^2 x = 0\).

Factor out \(e^{-5x} \tan x\):

\(e^{-5x} \tan x (-5 \tan x + 2 \sec^2 x) = 0\).

Since \(e^{-5x} \neq 0\) and \(\tan x = 0\) gives \(x = 0\), we solve:

\(-5 \tan x + 2 \sec^2 x = 0\).

Using \(\sec^2 x = 1 + \tan^2 x\), we get:

\(-5 \tan x + 2(1 + \tan^2 x) = 0\).

Simplifying, we have:

\(2 \tan^2 x - 5 \tan x + 2 = 0\).

Solving this quadratic equation in \(\tan x\), we find:

\(\tan x = 0.546\) or \(\tan x = 1.107\).

Thus, \(x = \arctan(0.546) \approx 0.464\) and \(x = \arctan(1.107) \approx 1.107\).

Therefore, the \(x\)-coordinates of the stationary points are \(x = 0, 0.464, 1.107\).