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Let $F(x-y,y-z,z-x)=0$,find $\frac{\partial z}{\partial x},\frac{\partial z}{\partial y}$.

This is a homework problem,I don't know how to do,appreciate any help.

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$\frac{\partial z}{\partial x}=0$. I think you wanted to write $\frac{\partial F}{\partial x}$, right? – Avitus Jun 19 '13 at 6:05
No...It is $\frac{\partial z}{\partial x}$...@Avitus – Luqing Ye Jun 19 '13 at 6:06
Could you please tell me why $\frac{\partial z}{\partial x}=0$?@Avitus – Luqing Ye Jun 19 '13 at 6:12
$z$ and $x$ are simply coordinates in $\mathbb R^{3}$. They satisfy $\frac{\partial z}{\partial x}=0$ because the function $z:\mathbb R^{3}\rightarrow \mathbb R$ $p\mapsto z(p):=p_z$, denoting by $p_z$ the $3$rd coordinate of any point $p$ in $\mathbb R^{3}$ does not depend on $x$. – Avitus Jun 19 '13 at 6:21
Thanks...It clarifies my concept...@Avitus – Luqing Ye Jun 19 '13 at 6:25

Let's have parameters such that $p=x-y$, $r=y-z$ and $s=z-x$ and we can say that $$F\big(p(x,y),r(y,z),s(z,x)\big)=0$$ it follows that $$dF=\frac{\partial F}{\partial p}dp+\frac{\partial F}{\partial r}dr+\frac{\partial F}{\partial s}ds=0$$ By using differential forms for parameters $$dF=\frac{\partial F}{\partial p}(dx-dy)+\frac{\partial F}{\partial r}(dy-dz)+\frac{\partial F}{\partial s}(dz-dx)=0$$ $$dF=\bigg(\frac{\partial F}{\partial p}-\frac{\partial F}{\partial s}\bigg)dx+\bigg(\frac{\partial F}{\partial r}-\frac{\partial F}{\partial p}\bigg)dy+\bigg(\frac{\partial F}{\partial s}-\frac{\partial F}{\partial r}\bigg)dz=0$$ Partial derivative of $z$ wrt $x$ means that you keep $y$ fixed or in other terms $dy=0$ $$\frac{\partial z}{\partial x}=-\frac{\frac{\partial F}{\partial p}-\frac{\partial F}{\partial s}}{\frac{\partial F}{\partial s}-\frac{\partial F}{\partial r}}$$ and if $dx=0$ $$\frac{\partial z}{\partial y}=-\frac{\frac{\partial F}{\partial r}-\frac{\partial F}{\partial p}}{\frac{\partial F}{\partial s}-\frac{\partial F}{\partial r}}$$

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