First, separate the variables:

\(\sin^2 3y \, dy = 4 \sec 2x \tan 2x \, dx\).

Integrate both sides:

\(\int \sin^2 3y \, dy = \int 4 \sec 2x \tan 2x \, dx\).

The left side integrates to:

\(\frac{1}{2}y - \frac{1}{12}\sin 6y\).

The right side integrates to:

\(2 \sec 2x\).

Using the initial condition \(y = 0\) when \(x = \frac{1}{6}\pi\), find the constant of integration:

\(\frac{1}{2}(0) - \frac{1}{12}\sin(0) = 2 \sec \left(\frac{1}{3}\pi\right) - C\).

\(C = 2 \sec \left(\frac{1}{3}\pi\right)\).

Substitute back to find \(x\) when \(y = \frac{1}{6}\pi\):

\(\frac{1}{2}\left(\frac{1}{6}\pi\right) - \frac{1}{12}\sin\left(\frac{1}{2}\pi\right) = 2 \sec 2x - 2 \sec \left(\frac{1}{3}\pi\right)\).

Solve for \(x\):

\(x = 0.541\).