First, separate the variables:

\(e^{4x} \frac{dy}{dx} = \cos^2 3y\)

\(\frac{dy}{\cos^2 3y} = e^{-4x} dx\)

Integrate both sides:

\(\int \sec^2 3y \, dy = \int e^{-4x} \, dx\)

\(\frac{1}{3} \tan 3y = -\frac{1}{4} e^{-4x} + C\)

Use the initial condition \(y = 0\) when \(x = 2\):

\(\frac{1}{3} \tan(0) = -\frac{1}{4} e^{-8} + C\)

\(0 = -\frac{1}{4} e^{-8} + C\)

\(C = \frac{1}{4} e^{-8}\)

Substitute back to find \(y\):

\(\frac{1}{3} \tan 3y = -\frac{1}{4} e^{-4x} + \frac{1}{4} e^{-8}\)

\(\frac{1}{3} \tan 3y = \frac{1}{4} e^{-8} - \frac{1}{4} e^{-4x}\)

\(\tan 3y = \frac{3}{4} e^{-8} - \frac{3}{4} e^{-4x}\)

\(y = \frac{1}{3} \arctan \left( \frac{3}{4} e^{-8} - \frac{3}{4} e^{-4x} \right)\)