x^2+y^2+z^2=1分别对x, 及y求导,得:2x+2z z'(y)=0--> z'(x)= -x/z2y+2z z'(y)=0--->z'(y)=-y/z因此有:Ux=yz'(x)+z'(x) x+z+y=-yx/z-x^2/z+z+y=y+z-x(y+x)/zUxy=1+z'(y)-x[ z-(y+x)z'(y)]/z^2=1-y/z-x[z+(y+x)y/z]/z^2=1-yz-x(z^2+y+x)/z^3