>>363
(xy)' を計算してみればわかる
(xy)' = x'y + xy' = 1y + xy' = y + xy'

3xy'+2y=1-x^2 みたいだったら両辺 x^(-1/3) 掛けるんだろうな
3x^(2/3)y'+2x^(-1/3)y = (1-x^2)x^(-1/3)
3(x^(2/3)y)' = (1-x^2)x^(-1/3)