本题x-y是变量,只能求最值
∵tanx=3tany
∴tan(x-y)
=(tanx-tany)/(1+tanxtany)
=(3tany-tany)/(1+3tan²y)
=2tany/(1+3tan²y)
∵0＜y≤x＜∏/2
∴tany>0
∴1+3tan²y≥2√3*tany
∴2tany/(1+3tan²y)≤√3/3
（当tan²y=1/3,tany=√3/3时取等号）
∴tan(x-y)≤√3/3
∴x-y≤π/6
即x-y的最大值为π/6