两边对x求导:
y+xy'-e^x+e^yy'=0
(x-e^y)y'=e^x-y
y'=(e^x-y)/(x-e^y)
将x=0代入原方程,得:-1+e^y=0,得:y(0)=0
因此y'(0)=(e^0-0)/(0-e^0)=-1