1:f'(x)=(1-lnx)/x^2
f'(1/e)=2*e^2
y=f(1/e)=-e
函数y=f(x)的图象在x=1/e处的切线方程 L
L:Y=2*e^2(X-1/e)-e
2:f'(x)=(1-lnx)/x^2
令f'(x)=0
得出x=e
y=f(x)的最大值 =f(e)=1/e