To solve the differential equation \(\frac{dy}{dx} = e^{3y} \sin^2 2x\), we first separate the variables:

\(\frac{1}{e^{3y}} dy = \sin^2 2x \, dx\).

Integrate both sides:

\(\int \frac{1}{e^{3y}} \, dy = \int \sin^2 2x \, dx\).

The left side becomes \(-\frac{1}{3} e^{-3y}\).

For the right side, use the double angle formula: \(\sin^2 2x = \frac{1}{2} (1 - \cos 4x)\).

Integrate to get \(\frac{1}{2} \left( x - \frac{1}{4} \sin 4x \right)\).

Combine the results:

\(-\frac{1}{3} e^{-3y} = \frac{1}{2} \left( x - \frac{1}{4} \sin 4x \right) + C\).

Use the initial condition \(y = 0\) when \(x = 0\) to find \(C\):

\(-\frac{1}{3} = C\).

Substitute \(x = \frac{1}{2}\) to find \(y\):

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

Simplify to find \(y = 0.175\) or \(y = -\frac{1}{3} \ln \left( \frac{1}{4} + \frac{3}{8} \sin 2 \right)\).