x^2+4y^2-2xy+12y+12
=x^2-2xy+y^2+3(y^2+4y+4)
=(x-y)^2+3(y+2)^2
=0
所以：x-y=0,y-2=0
即x=y=-2
则x^y=(-2)^(-2)=1/(-2)^2=1/4