???$解:由題意得4x^2-4x+1+y^2-8y+16=0,即(2x-1)^2+(y-4)^2=0$???
???$∴2x-1=0,y-4=0即x=\frac 12,y=4$???
???$原式=\frac 23x×3\sqrt{x}-4x×\frac {\sqrt{xy}}x=2x\sqrt{x}-4\sqrt{xy}$???
???$=2×\frac {1}{2}×\sqrt{\frac {1}{2}}-4×\sqrt{\frac {1}{2}×4}$
$=\frac {\sqrt{2}}{2}-4\sqrt{2}$???
???$=-\frac {7\sqrt{2}}{2}$???