两端对x求偏导：1+z'x=yf'(x^2-z^2)(2x-2zz'x)[1+2yzf'(x^2-z^2)]z'x=2xyf'(x^2-z^2)-1z'x=[2xyf'(x^2-z^2)-1]/[1+2yzf'(x^2-z^2)]两端对y求偏导：z'y=f(x^2-z^2)+yf'(x^2-z^2)(-2zz'y)[1+2yzf'(x^2-z^2)]z'y=f(x^2-...可你知道吗，答案等于x，就是如此的简单，WHY？再化一步就是了，利用x+z=yf(x^2-z^2)，[yf(x^2-z^2)+2xyzf'(x^2-z^2)-z]/[1+2yzf'(x^2-z^2)]=[x+z+2xyzf'(x^2-z^2)-z]/[1+2yzf'(x^2-z^2)]=x