y-xe^y=1
两边同时对x求导：y'-(e^y+xy'e^y)=0
∴y'=e^y/(1-xe^y)
当x=0时,y=1,代入上式,那么y'|(x=0)=e/(1-0)=e
所以切线方程是：y-1=e(x-0),即ex-y+1=0