设方程xyz-xz-yz=1确定函数z=ƒ(x，y)，求∂z/∂x,∂z/∂y.
设F(x，y，z)=xyz-xz-yz-1，F′x=yz-z，F′y=xz-z，F′z=xy-x-y，所以∂z/∂x=-(F′x/F′z)=-[(yz-z)/(xy-x-y)]=[z(1-y)]/(xy-x-y)，∂z/∂y=-(F′x/F′z)=-[(xz-z)/(xy-x-y)]=[z(1-y)]/(xy-x-y)
设方程xyz-xz-yz=1确定函数z=ƒ(x，y)，求∂z/∂x,∂z/∂y.
设F(x，y，z)=xyz-xz-yz-1，F′x=yz-z，F′y=xz-z，F′z=xy-x-y，所以∂z/∂x=-(F′x/F′z)=-[(yz-z)/(xy-x-y)]=[z(1-y)]/(xy-x-y)，∂z/∂y=-(F′x/F′z)=-[(xz-z)/(xy-x-y)]=[z(1-y)]/(xy-x-y)
