$z=z(x,y)$ is a positive differentiable function defined on $W=\{(x,y): x,y >0\}$, such that $F(x+\frac{z}{y}, y+\frac{z}{x}) = 0$, where $F:\mathbb R^2 \to \mathbb R$ and $\nabla F \neq 0$.
Show that $x\frac{\partial z}{\partial x} + y\frac{\partial z}{\partial y} = z-xy$
What I did:
I derived using the chain rule and hoped to show what we wanted. Derive $F(x+\frac{z}{y}, y+\frac{z}{x}) = 0$ wrt both variables. First $x$:
$\frac{\partial F}{\partial x}\frac{\partial(x+\frac{z}{y})}{\partial x} + \frac{\partial F}{\partial y}\frac{\partial(y+\frac{z}{x})}{\partial x} = \frac{\partial F}{\partial x}(1+\frac{1}{y}\frac{\partial z}{\partial x})+\frac{\partial F}{\partial y}\frac{x\frac{\partial z}{\partial x}-z}{x^2} = 0$
and $y$ similarly $\frac{\partial F}{\partial x}\frac{y\frac{\partial z}{\partial y}-z}{y^2} + \frac{\partial F}{\partial y}(1+\frac{1}{x}\frac{\partial z}{\partial y})=0$
I was hoping by adding or subtracting these equations something cancels out (most importantly the derivatives of $F$), but no dice.
How do we continue? why is it important that $z$ is positive and what $W$ is?
The chain rule says that $$ 0=\nabla F\!\left(x+\frac zy,y+\frac zx\right)\cdot\left[\left(1+\frac{z_x}y,\frac{z_x}x-\frac{z}{x^2}\right)\mathrm{d}x+\left(\frac{z_y}y-\frac{z}{y^2},1+\frac{z_y}x\right)\mathrm{d}y\right] $$ for any $\mathrm{d}x$ and $\mathrm{d}y$
This means that both $$ \left(1+\frac{z_x}y,\frac{z_x}x-\frac{z}{x^2}\right) $$ and $$ \left(\frac{z_y}y-\frac{z}{y^2},1+\frac{z_y}x\right) $$ are perpendicular to $$ \nabla F\!\left(x+\frac zy,y+\frac zx\right) $$ and are thus, parallel. This means that $$ \left(1+\frac{z_x}y\right)\left(1+\frac{z_y}x\right) =\left(\frac{z_x}x-\frac{z}{x^2}\right)\left(\frac{z_y}y-\frac{z}{y^2}\right)\\ \Downarrow\\ 1+\frac{z_x}y+\frac{z_y}x =\left(-\frac{z_x}{y}-\frac{z_y}{x}+\frac{z}{xy}\right)\frac{z}{xy}\\ \Downarrow\\ \left(\frac{z}{xy}+1\right)\left(\frac{z_x}y+\frac{z_y}x\right)=\frac{z^2}{x^2y^2}-1\\ \Downarrow\\ \frac{z_x}y+\frac{z_y}x=\frac{z}{xy}-1\\ \Downarrow\\ x\,z_x+y\,z_y=z-xy $$