Separate the variables: \(\frac{1}{\cos^2 y} dy = 4 \tan x \, dx\).

Integrate both sides: \(\int \sec^2 y \, dy = \int 4 \tan x \, dx\).

The integrals are: \(\tan y = 4 \ln |\sec x| + C\).

Use the initial condition \(x = 0\) and \(y = \frac{1}{4}\pi\):

\(\tan \left( \frac{1}{4}\pi \right) = 4 \ln |\sec 0| + C\).

\(1 = 4 \cdot 0 + C \Rightarrow C = 1\).

The solution is \(\tan y = 4 \ln |\sec x| + 1\).

Substitute \(y = \frac{1}{3}\pi\):

\(\tan \left( \frac{1}{3}\pi \right) = 4 \ln |\sec x| + 1\).

\(\sqrt{3} = 4 \ln |\sec x| + 1\).

\(4 \ln |\sec x| = \sqrt{3} - 1\).

\(\ln |\sec x| = \frac{\sqrt{3} - 1}{4}\).

\(|\sec x| = e^{\frac{\sqrt{3} - 1}{4}}\).

\(\sec x = e^{\frac{\sqrt{3} - 1}{4}}\).

\(x = \arccos \left( \frac{1}{e^{\frac{\sqrt{3} - 1}{4}}} \right)\).

\(x \approx 0.587\).