xy-e^x+e^y=0
对x求导
则(xy)'=1*y+x*y'
(e^x)'=e^x
(e^y)=e^y*y'
所以y-e^x+(x+e^y)y'=0
y'=(e^x-y)/(x+e^y)
所以dy/dx=(e^x-y)/(x+e^y)