x*e^y-y+e=0对x求导，可知e^y+xe^y*y'-y'+0=0则y'=e^y/(1-xe^y)再求二阶导可知y''=[e^y/(1-xe^y)]'=[e^y*y'(1-xe^y)+e^y(e^y+xe^y*y')]/(1-xe^y)]^2=[e^(2y)+e^(2y)*(1+xe^y/(1-xe^y)]/(1-xe^y)]^2=[2e^(2y)-xe^(3y)]/(1-xe^y)]^3
