f'(x)=1/x²-lnx/x²-1,
当x=1时,f'(x)=0,
当0＜x＜1时,1/x²-1＞0,-lnx/x²＞0,
f'(x)=(1/x²-1)-lnx/x²＞0,
当x＞1时,f'(x)＜0,
所以f(x)在(0,+∞)上先是递增,再递减,
且在x=1时,取到最大值-1.