6（1）两端求微分，得 (e^y)dy+ydx+xdy = 0，解得 y' = dy/dx = -y/[(e^y)+x]，于是 y" = dy'/dx = (d/dx){-y/[(e^y)+x]} = -{y'[(e^y)+x]-y [(e^y)*y'+1]}/[(e^y)+x]² = ……
