To solve the differential equation \(\frac{dy}{dx} = e^{-2y} \tan^2 x\), we first separate variables:

\(e^{2y} \, dy = \tan^2 x \, dx\).

Integrate both sides:

\(\int e^{2y} \, dy = \int \tan^2 x \, dx\).

The left side integrates to \(\frac{1}{2} e^{2y}\), and the right side integrates to \(\tan x - x + C\), where \(C\) is the constant of integration.

Thus, \(\frac{1}{2} e^{2y} = \tan x - x + C\).

Using the initial condition \(y = 0\) when \(x = 0\), we find:

\(\frac{1}{2} e^{0} = \tan 0 - 0 + C\).

This gives \(\frac{1}{2} = C\).

Substitute \(C = \frac{1}{2}\) back into the equation:

\(\frac{1}{2} e^{2y} = \tan x - x + \frac{1}{2}\).

Now, solve for \(y\) when \(x = \frac{1}{4}\pi\):

\(\frac{1}{2} e^{2y} = \tan \frac{1}{4}\pi - \frac{1}{4}\pi + \frac{1}{2}\).

Calculate \(\tan \frac{1}{4}\pi = 1\), so:

\(\frac{1}{2} e^{2y} = 1 - \frac{1}{4}\pi + \frac{1}{2}\).

\(\frac{1}{2} e^{2y} = \frac{3}{2} - \frac{1}{4}\pi\).

\(e^{2y} = 3 - \frac{1}{2}\pi\).

\(2y = \ln(3 - \frac{1}{2}\pi)\).

\(y = \frac{1}{2} \ln(3 - \frac{1}{2}\pi)\).

Calculate \(y \approx 0.179\).