Assume the equation $f(\frac{x}{z}, \frac{y}{z})=0$ defines an implicit function $z=g(x,y)$, and that all functions are differentiable. Show that $$x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=z$$
My attempt: Define $u(x,y)=\frac{x}{z}, v(x,y)=\frac{y}{z}$. Then
$$f'_x=\frac{f'_v}{z}-\frac{z'_y}{z^2}(f'_u x + f'_v y)$$ $$f'_y=\frac{f'_u}{z}-\frac{z'_x}{z^2}(f'_u x + f'_v y)$$
I tried to isolate $z'_x$ and $z'_y$ but that didn't yield anything.
Thanks in advance for any help.
On
We have $\partial_{x}\left[f\left(\frac{x}{g\left(x,y\right)},\,\frac{y}{g\left(x,y\right)}\right)\right]=\partial_{1}f\left(\frac{x}{g\left(x,y\right)},\,\frac{y}{g\left(x,y\right)}\right)\cdot\left(\frac{1}{g\left(x,y\right)}-x\frac{\partial_{x}g\left(x,y\right)}{\left(g\left(x,y\right)\right)^{2}}\right)-\partial_{2}f\left(\frac{x}{g\left(x,y\right)},\,\frac{y}{g\left(x,y\right)}\right)\cdot y\frac{\partial_{y}g\left(x,y\right)}{\left(g\left(x,y\right)\right)^{2}}=0$
and $\partial_{y}\left[f\left(\frac{x}{g\left(x,y\right)},\,\frac{y}{g\left(x,y\right)}\right)\right]=\partial_{2}f\left(\frac{x}{g\left(x,y\right)},\,\frac{y}{g\left(x,y\right)}\right)\cdot\left(\frac{1}{g\left(x,y\right)}-y\frac{\partial_{y}g\left(x,y\right)}{\left(g\left(x,y\right)\right)^{2}}\right)-\partial_{1}f\left(\frac{x}{g\left(x,y\right)},\,\frac{y}{g\left(x,y\right)}\right)\cdot x\frac{\partial_{x}g\left(x,y\right)}{\left(g\left(x,y\right)\right)^{2}}=0$
Thus subtracting the above two equations: $\left(\partial_{1}f\left(\frac{x}{g\left(x,y\right)},\,\frac{y}{g\left(x,y\right)}\right)-\partial_{2}f\left(\frac{x}{g\left(x,y\right)},\,\frac{y}{g\left(x,y\right)}\right)\right)\cdot\frac{1}{g\left(x,y\right)}=0$
Thus writing $c\left(x,\, y\right)=\partial_{1}f\left(\frac{x}{g\left(x,y\right)},\,\frac{y}{g\left(x,y\right)}\right)=\partial_{2}f\left(\frac{x}{g\left(x,y\right)},\,\frac{y}{g\left(x,y\right)}\right)$ we have $$c\left(x,y\right)\cdot\left(\frac{1}{g\left(x,y\right)}-x\frac{\partial_{x}g\left(x,y\right)}{\left(g\left(x,y\right)\right)^{2}}-y\frac{\partial_{y}g\left(x,y\right)}{\left(g\left(x,y\right)\right)^{2}}\right)=0$$ $$c\left(x,y\right)\cdot\left(g\left(x,y\right)-x\partial_{x}g\left(x,y\right)-y\partial_{y}g\left(x,y\right)\right)=0$$ and thus: $$g\left(x,y\right)-x\partial_{x}g\left(x,y\right)-y\partial_{y}g\left(x,y\right)=0$$
qed
So we write $F(x,y,z) = f(x/z,y/z)$. If we assume that $\partial F/\partial z\ne 0$ on $F=0$, then by the implicit function theorem this equation defines $z$ locally as a differentiable function $g$ of $x$ and $y$. Morever, we have the formula $$\frac{\partial g}{\partial x} = -\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial z}},$$ and similarly for the $y$ partial. But by the chain rule, writing $f=f(u,v)$, we have $$\frac{\partial F}{\partial x} = \frac{\partial f}{\partial u}(x/z,y/z)\cdot \frac 1z \qquad\text{and}\qquad \frac{\partial F}{\partial z} = \frac{\partial f}{\partial u}(x/z,y/z)\cdot \left(-\frac x{z^2}\right)+ \frac{\partial f}{\partial v}(x/z,y/z)\cdot \left(-\frac y{z^2}\right).$$ Similarly, $$\frac{\partial F}{\partial y} = \frac{\partial f}{\partial v}(x/z,y/z)\cdot \frac1z.$$ Then (with $z=g(x,y)$ inserted throughout) $$x\frac{\partial g}{\partial y} + y\frac{\partial g}{\partial y} = -\frac{\frac xz \frac{\partial f}{\partial u} + \frac yz\frac{\partial f}{\partial v}}{\frac{\partial f}{\partial u}\cdot \left(-\frac x{z^2}\right)+ \frac{\partial f}{\partial v}\cdot \left(-\frac y{z^2}\right)}=z=g(x,y),$$ as desired.