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If $F(x/y,z/x)=0$ we want to show that $$xz_x+yz_y=0.$$ I am not sure if I can write $z_x=F_x/F_z$ just to use the assumption? Thank you friends.

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  • $\begingroup$ No more condition? $\endgroup$ – Shuchang Aug 17 '14 at 15:54
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Nancy

It seems the conclusion isnot right and must change to $~xz_{x}+yz_{y}=z~$. For this, define $u=\frac{y}{x}$ and $v=\frac{z}{x}$. As you mentioned $z$ is a differentiable function of $x$ and $y$. Now use the chaine rule. That is , $\frac{\partial{F}}{\partial{u}}\frac{\partial{}u}{\partial{x}}+\frac{\partial{F}}{\partial{v}}\frac{\partial{v}}{\partial{x}}=0~$ and $~\frac{\partial{F}}{\partial{u}}\frac{\partial{u}}{\partial{y}}+\frac{\partial{F}}{\partial{v}}\frac{\partial{v}}{\partial{y}}=0.~$These equations yield $F_{u}(\frac{-y}{x^2})+F_{v}(\frac{xz_{x}-z}{x^2})=0~$and $~F_{u}(\frac{1}{x})+F_{v}(\frac{z_{y}}{x})=0~.$ Separating and then Dividing the last two equations we obtain, $xz_{x}+yz_{y}=z.$

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