Let $F(x,y,z) = f(x/y, y/z, z/x)$ and show that i've been strugling in a calculus problem. Let:
$$F(x,y,z) = f(\frac{x}{y}, \frac{y}{z}, \frac{z}{x})$$
show that $$ x\frac{∂F}{∂x} + y\frac{∂F}{∂y} + z\frac{∂F}{∂z} = 0.$$
In my solution i assume that : $$(I)(\frac{∂F}{∂x}, \frac{∂F}{∂y}, \frac{∂F}{∂z}) = (\frac{∂f}{∂x}, \frac{∂f}{∂y}, \frac{∂f}{∂z}).$$ And i let $ u = \frac{x}{y}, v = \frac{y}{z}, w = \frac{z}{x}$
then i calculate the partial derivatives : $$\frac{∂f}{∂x} = \frac{∂f}{∂u}. \frac{1}{y} - \frac{∂f}{∂w}. \frac{z}{x^2}$$
$$\frac{∂f}{∂y} = \frac{∂f}{∂u}. \frac{-x}{y^2} + \frac{∂f}{∂v}. \frac{1}{z}$$
$$\frac{∂f}{∂z} = \frac{∂f}{∂v}. \frac{-y}{z^2} + \frac{∂f}{∂w}. \frac{1}{x}$$
And did the inner product in both sides of (I) :
$$<(\frac{∂F}{∂x}, \frac{∂F}{∂y}, \frac{∂F}{∂z}),(x,y,z)> = <(\frac{∂f}{∂x}, \frac{∂f}{∂y}, \frac{∂f}{∂z}),(x,y,z)> =0.$$
That implies:
$$ x\frac{∂F}{∂x} + y\frac{∂F}{∂y} + z\frac{∂F}{∂z} = 0.$$
But i'm not sure if i can assume (I).
Thanks.
 A: More simply assuming $u=x/y$, $v=y/z$, $w= z/x$, we have
$$\frac{∂F}{∂x} =\frac1yf_{u}-\frac z{x^2} f_{w}$$
$$\frac{∂F}{∂y} =\frac1zf_{v}-\frac x{y^2} f_{u}$$
$$\frac{∂F}{∂z} =\frac1xf_{w}-\frac y{z^2} f_{v}$$
and therefore
$$x\frac{∂F}{∂x} + y\frac{∂F}{∂y} + z\frac{∂F}{∂z} = 0$$
A: I think it's clearer to express $F$ as a composition $f\circ g$ where $g:\mathbb R^3\to \mathbb R^3:(x,y,z)\mapsto (x/y,y/z,z/x)$. Then, $F=f\circ g:\mathbb R^3\to \mathbb R$ and the derivatives may be read off the Jacobian matrix of the composition:
Fix $(x,y,z)\in \mathbb R^3.$ Then, $DF((x,y,z))=DF(g(x,y,z))\circ Dg(x,y,z)$. The matrix elements of this linear transformation with respect to the standard basis are
$\partial F_1(x,y,z)=\partial_1f(g((x,y,z)))(1/y)+\partial_2f(g((x,y,z)))(0)+\partial_3f(g((x,y,z))(-z/x^2))$
$\partial F_2(x,y,z)=\partial_1f(g((x,y,z)))(-x/y^2)+\partial_2f(g(x,y,z)))(1/z)+\partial_3f(g(x,y,z)))(0)$
$\partial F_3(x,y,z)=\partial_1f(g(x,y,z)))(0)+\partial_2f(g(x,y,z)))(-y/z^2)+\partial_3f(g(x,y,z))(1/x))$
Using these data, you can finish the proof. 
A: $\partial_x F={1\over y}\partial_x f(x/y,y/z,z/x)-{z\over x^2}\partial_zf(x/y,y/z,z/x)$
$\partial_y F={1\over z}\partial_y f(x/y,y/z,z/x)-{x\over y^2}\partial_xf(x/y,y/z,z/x)$
$\partial_z F={1\over x}\partial_z f(x/y,y/z,z/x)-{y\over z^2}\partial_yf(x/y,y/z,zx)$
