F(x,y,z)=0,z=z(x,y),为什么F对x的偏导数仍然是x,y,z的函数.不是应该只是x的函数吗

F(x/z,y/z)=0,求dz过程 (x+y-z)(x-y+z)= 分解因式:f(x,y,z)=x^2(y-z)+y^2(z-x)+z^2(x-y) 已知 (x+y-z)/z=(x-y+z)/y=(y+z-x)/x,且xyz≠0,求代数式 ((x+y)(y+z)(x+z))/xyz x分之y+z=y分之z+x=z分之x+y(x+y+z不等于0）,求x+y+z分之x+y-z 分式加减法：已知x+y/z=x+z/y=y+z/x(x+y+z≠0）,求x+y-z/x+y+z 已知x+y/z=x+z/y=y+z/x(x+y+z≠0）,求x+y-z/x+y+z的步骤 已知x+y/z=x+z/y=y+z/x(x+y+z≠0）,求x+y-z/x+y+z的步骤 已知：x^2/(z+y)+y^2/(x+z)+z^2/(x+y)=0,求x/(z+y)+y/(x+z)+z/(x+y)的值. x^2/(z+y)+y^2/(x+z)+z^2/(x+y)=0,求x/(z+y)+y/(x+z)+z/(x+y)的值 X+Y+Z=? x/y=(x+z)/(y+z)y/z=(x+y)/(x+z) 设f(x,y,z)=e^x*y*z^2,其中z=z(x,y)是由x+y=z+x*e^(z-x-y)确定的隐函数,则f'x(0,1,1)= 已知 x/(y+z)+y/(z+x)+z/(x+y)=1求 (x*x)/(y+z)+(y*y)/(x+z)+(z*z)/(x+y)=? z=f(u) u=x/y,求x*∂z/∂x +y*z∂z/∂y f(x,y,z)=In{[x^α+z^(1-α-β)]/y^β}+x/z+(z/x)^β,求 ∂f(x,y,z)/∂y (y+z)/x=(z+x)/y=(x+y)/z求x+y-z/x+y+z的值 已知x+y+z=0求证x*x*x+y*y*y+z*z*z=3xyz