Separation gives \((1+y)e^{-3y}\,dy=d\theta/(1+\cos2\theta)=\frac12\sec^2\theta\,d\theta\). Integrating by parts on the left: \[-\frac13(1+y)e^{-3y}-\frac19e^{-3y}=\frac12\tan\theta+C.\] Applying the initial condition gives \(C=-17/18\). Therefore \[\tan\theta=\frac{17}{9}-\frac{2(3y+4)}9e^{-3y}.\] At y=1, \(\tan\theta=\frac{17}{9}-\frac{14}{9}e^{-3}\).