Homogenous PDE, changing of variable I have an PDE $\dfrac{df}{d \xi}-\xi\dfrac{d x_1}{d \xi}=0$ homogeneous for $\xi$, where $f:\mathbb{R}^n\to\mathbb{R}$ is function of $x_1,\cdots,x_n:\mathbb{R}\to\mathbb{R}$, which in turn are functions of $\xi$ (so, $\dfrac{df}{d \xi}$ is a total derivative).
I also have $\xi=y/z\in\mathbb{R}$, where $y,z\in\mathbb{R}$, and one said that once the PDE is homogeneous in $\xi$ I have $\dfrac{df}{d y}-\xi\dfrac{d x_1}{d y}=0$.
I missed this gap. I imagine that I can multiply the first ODE by $\dfrac{d\xi}{d y}=\dfrac{1}{z}$, so
$$\dfrac{df}{d \xi}\dfrac{d\xi}{d y}-\xi\dfrac{d x_1}{d \xi}\dfrac{d\xi}{d y}=\sum_{i=1}^n \dfrac{\partial f}{\partial x_i}\dfrac{d x_i}{d\xi}\dfrac{d\xi}{d y}-\xi\dfrac{d x_1}{d y}=\sum_{i=1}^n \dfrac{\partial f}{\partial x_i}\dfrac{d x_i}{d y}-\xi\dfrac{d x_1}{d y}=\dfrac{d f}{d y}-\xi\dfrac{d x_1}{d y}=0.$$
But I am not sure. Indeed, I'd like to know how state this only from the fact of homogeneous.
Thanks so much.

Related:
Equality between two total derivatives

 A: Assume that $f:\textbf{R}^n\rightarrow\textbf{R}$ and $f=f(x_1,x_2,\ldots,x_n)$ is a function of $n$ variables. By saying that $x_i=x_i(\xi)$, then $C:\overline{x}=\{x_1(\xi),x_2(\xi),\ldots,x_n(\xi)\}$, $\xi\in\textbf{R}$, then $C$ is one dimentional object in $\textbf{R}^n$ and  hence $C$ is a curve of $\textbf{R}^n$. Then
$$
\frac{df}{d\xi}=\sum^{n}_{k=1}\frac{\partial f}{\partial x_k}\frac{dx_k}{d\xi}
$$
is the derivative of $f$ allong $C$ (or total derivative of $f$ allong the curve $C$). You also have the equation:
$$
\frac{df}{d\xi}-\xi\frac{dx_1}{d\xi}=0\Leftrightarrow \sum^{n}_{k=1}\frac{\partial f}{\partial x_k}\frac{dx_k}{d\xi}=\xi\frac{dx_1}{d\xi} \tag 1
$$
If $\xi=u y$, then $\frac{d\xi}{dy}=u$. Hence
$$
\frac{df}{d\xi}-\xi\frac{dx_1}{d\xi}=0\Leftrightarrow \frac{df}{dy}\frac{dy}{d\xi}-\xi\frac{dx_1}{dy}\frac{dy}{d\xi}=0\Leftrightarrow \frac{df}{dy}\frac{1}{u}-\xi\frac{dx_1}{dy}\frac{1}{u}=0\Leftrightarrow
$$
$$
\frac{df}{dy}-\xi\frac{dx_1}{dy}=0.\tag 2
$$
This answer your first question about the change of variables.
About the homogenicity
However if $f$ is homogeneous function then we have even more
If the function $f$ is homogeneous of degree $\lambda$. Then setting $x_i=uy_i$ in equation (1) we have, (knowing that $f(x_1,x_2,\ldots,x_n)$ and $(x_1,x_2,\ldots,x_n)\rightarrow x_1$ are homogeneous i.e.   $f(uy_1,uy_2,\ldots,uy_n)=u^{\lambda}f(y_1,y_2,\ldots,y_n)$ and $(ux_1)=ux_1$ of degree 1):
$$
\sum^{n}_{k=1}\frac{\partial f}{\partial x_k}(uy_1,uy_2,\ldots,uy_n)\left(u\frac{dy_k}{d\xi}\right)-\xi\left(u\frac{dy_1}{d\xi}\right)=0\Leftrightarrow
$$
$$
u^{\lambda-1}\sum^{n}_{k=1}\frac{\partial f}{\partial y_k}(y_1,y_2,\ldots,y_n)\left(u\frac{dy_k}{d\xi}\right)-\xi u\frac{dy_1}{d\xi}=0
$$
$$
u^{\lambda-1}\sum^{n}_{k=1}\frac{\partial f}{\partial y_k}(y_1,y_2,\ldots,y_n)\frac{dy_k}{d\xi}-\xi \frac{dy_1}{d\xi}=0.\tag 3
$$
(That is because when $f(x_1,x_2,\ldots ,x_n)$ is homogeneous of degree $\lambda$, then $\frac{\partial f}{\partial x_{j}}$ is homogeneous of degree $\lambda-1$ i.e. $\frac{\partial f}{\partial x_j}(uy_1,uy_2,\ldots,uy_j,\ldots,uy_n)=u^{\lambda-1}\frac{\partial f}{\partial x_j}(y_1,y_2,\ldots,y_n)$). Hence when $\lambda=1$, then (3) becomes:
$$
\sum^{n}_{k=1}\frac{\partial f}{\partial y_k}\frac{dy_k}{d\xi}-\xi\frac{dy_1}{d\xi}=0.\tag 4
$$
Hence if $f=f(x_1,x_2,\ldots,x_n)$ is homogeneous of degree 1, then equation (1) is homogeneous PDE (invariant under any transformation of variables of the form $x_i=uy_i$, $i=1,2,\ldots,n$).
