z^2-xy+z=1,  求 ∂²z/∂x∂y
  等式兩邊對(duì)y求偏導(dǎo),得 2zz'<y>-x+z'<y>=0,得 z'<y>=x/(1+2z),
  等式兩邊對(duì)x求偏導(dǎo),得 2zz'<x>-y+z'<x>=0,  得 z'<x>=y/(1+2z),
  進(jìn)而得 z''<xy> = (1+2z-2yz'<y>)/(1+2z)^2 = [1+2z-2xy/(1+2z)]/(1+2z)^2
                        = [(1+2z)^2-2xy]/(1+2z)^3.
記 F=x+y-u-v=0,   G=x/y-sinu/sinv=0,  其中 u=u(x,y),v=v(x,y).
則 F'<x>=1-u'<x>-v'<x>=0,   G'<x>=1/y-(u'<x>cosusinv-v'<x>sinucosv)/(sinv)^2=0,
即  u'<x>+v'<x>=1
      u'<x>cosusinv-v'<x>sinucosv=(sinv)^2/y
當(dāng) sin(u+v)≠0 時(shí)聯(lián)立解得
u'<x>= [(sinv)^2/y+sinucosv]/sin(u+v),
        v'<x>=[-(sinv)^2/y+cosusinv]/sin(u+v).