Divide the equation by cos2y:
xsec2ydxdy−2tany=x3(2−x3).
Let v=tany, so sec2ydxdy=dxdv.
The equation becomes xdxdv−2v=x3(2−x3).
Dividing by x: dxdv−x2v=x2(2−x3).
Using integrating factor μ=x−2: dxd(x−2v)=2−x3.
Integrating: x−2v=2x−4x4+C.
From y(2)=0, we have tan0=0=16−16+4C, giving C=0.
Therefore tany=2x3−4x6.
At x=1: tan(y(1))=2−41=47.