解：y'=1/xy'(e)=1/e切线方程：y-1=1/e(x-e)ey-e=x-eey=xy=x/e发现方程，k'=-1/k=-1/1/e=-ey-1=-e(x-e)y-1=-ex+e^2ex+y-1-e^2=0答：切线方程和法线方程分别为y=x/e，ex+y-1-e^2=0。
