# Prove that the following differential equation holds

I'm given implicit function $z=F(\frac{x}{z},\frac{y}{z})$. For $z(x,y)$ I want to show that $$x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=z.$$ I'm using implicit function theorem. My reasoning goes as follows. I compute matrix of linear derivative of $\textbf{F}(x,y,z)=F(\frac{x}{z},\frac{y}{z})-z$ with respect to $z$, assume it's not singular and reverse it. Next I compute linear derivative of $\textbf{F}$ with respect to $x,y$. Then I multiply the last matrix by the reverse of linear derivative of $\textbf{F}$ (from left). Finally I get matrix of linear derivative of the function $z(x,y)$ which should easily yield the result. By computation I get $$D_{z}\textbf{F}=\frac{\partial F}{ \partial u} \frac{\partial u} {\partial z} + \frac{\partial F}{ \partial t} \frac{\partial t} {\partial z}-1=-x\frac{\partial F}{ \partial u} \frac{1} {z^{2}} -y \frac{\partial F}{ \partial t} \frac{1} {z^{2}}-1,$$ so the reverse of $D_{z}\textbf{F}$ is $$[D_{z}\textbf{F}]^{-1} = (\frac{\partial F}{ \partial u} \frac{\partial u} {\partial z} + \frac{\partial F}{ \partial t} \frac{\partial t} {\partial z}-1)^{-1}$$and partial derivatives of $z(x,y)$ came up to be $$\frac{\partial z}{ \partial x} = \frac{\frac{\partial F}{ \partial u} \frac{1} {z}}{[D_{x}\textbf{F}]^{-1}},$$ and $$\frac{\partial z}{ \partial y} = \frac{\frac{\partial F}{ \partial t} \frac{1} {z}}{[D_{x}\textbf{F}]^{-1}}.$$ So it's easy to see that the equation does not hold. My question is: is my reasoning correct? And did I messed up something with partial derivatives?

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