Separate: \(e^{-3y}\,dy=\sin^2(2x)\,dx\). Integration gives \(-\tfrac13e^{-3y}=x/2-\sin4x/8+C\). The initial values give \(C=-1/3\), so \(e^{-3y}=1-\tfrac32x+\tfrac38\sin4x\).

Thus \(y=-\tfrac13\ln(1-\tfrac32x+\tfrac38\sin4x)\), on the branch containing the initial point. At \(x=1/2\), \(\boxed{y=-\tfrac13\ln(\tfrac14+\tfrac38\sin2)=0.175}\) (3 s.f.).