xy=e^(x+y)两边对x求导,得：y+xy’=(1+y’)e^(x+y)移项,得：[x-e^(x+y)]y’=e^(x+y)-y整理得：y’=[e^(x+y)-y]/[x-e^(x+y)]将xy=e^(x+y)代入,即把e^(x+y)换成xy,得：y’=(xy-y)/(x-xy)所以dy/dx=(xy-y)/(x-xy)...