∵(x+1)²+|y-1|=0
∴x+1=0,y-1=0
∴x=-1,y=1
∴2(xy-5xy²)-(3xy²)-(3xy²-xy)
=2xy-10xy²-3xy²-3xy²+xy
=3xy-16xy²
=xy(3-16y)
=(-1)×1×(3-16×1)
=(-1)×1×(-13)
=13