xy=e^(x+y)(y+xy')=e^(x+y)*(x+y)'y+xy'=e^(x+y)(1+y')y+xy'=e^(x+y)+e^(x+y)(1+y')所以：dy/dx=y'=[e^(x+y)-y]/[x-e^(x+y)].
两边对x求导得y+xy'=(1+y')*e^(x+y)∴y'=[y-e^(x+y)]/[e^(x+y) -x]
