设x+y=a,x-y=b.则x=(a+b)/2,y=(a-b)/2
∵f(x+y,x-y)=xy+y²
∴f(a,b)=[(a+b)/2][(a-b)/2]+[(a-b)/2]²
=[(a-b)/2][(a+b)/2+(a-b)/2]
=a(a-b)/2
故f(x,y)=x(x-y)/2
设x+y=a,x-y=b.则x=(a+b)/2,y=(a-b)/2
∵f(x+y,x-y)=xy+y²
∴f(a,b)=[(a+b)/2][(a-b)/2]+[(a-b)/2]²
=[(a-b)/2][(a+b)/2+(a-b)/2]
=a(a-b)/2
故f(x,y)=x(x-y)/2