答案:解:∵x y z=1①,x2 y2 z2=3②∴①2﹣②可得:xy yz xz=﹣1∴xy z(x y)=﹣1∵x y z=1,∴x y=1﹣z∴xy=﹣1﹣z(x y)=﹣1﹣z(1﹣z)=z2﹣z﹣1∵x2 y2=3﹣z2≥2xy=2(z2﹣z﹣1)⇒3z2﹣2z﹣5≤0⇒﹣1≤z≤令f(z)=xyz=z3﹣z2﹣z,则f′(z)=3z2﹣2z﹣1=(z﹣1)(3z 1)令f′(z)>0,可得z>1或z<,∴f(z)在区间[﹣1,﹣]单调递增,在[﹣,1]单调递减,在[1,]单调递增,当z=﹣时,xyz的值为,当z=时,xyz的值为,∴xyz的最大值为.故答案为:.