Eliminate the arbitrary funcion - PDE first order I'm heading the book Elements Of Partial Differential Equations -Sneddon 1957. At chapter two exists this exercise
"Eliminate the arbitrary function $f$ fron the equation
$$ z= f\left(\dfrac{xy}{z}\right). $$
Anyone can help me?
Tks!
 A: Your question is incomplete, one need to read page 46 of
Sneddon's book to figure out what the problem really want.
To summarize, what the book want is start from an equation of the form
$$z = f(u)\quad\text{ where }\quad u = \frac{xy}{z}$$
derive a PDE for $z$ which doesn't involve the function $f(u)$ explicitly.
The tool you need is chain rule
for partial derivatives. 
Let $\;\displaystyle\;p = \frac{\partial z}{\partial x}\;$ and $\displaystyle\;q = \frac{\partial z}{\partial y}\;$ and taking partial derivatives of $z = f(u)$, we have
$$\begin{align}
\color{red}{p} \;&=
\frac{\partial f(u)}{\partial x}
= f'(u)\frac{\partial u}{\partial x}
= f'(u)\color{red}{\left(\frac{y}{z} - \frac{xy}{z^2}p\right)}\\
\color{blue}{q} \;&=
\frac{\partial f(u)}{\partial y}
= f'(u)\frac{\partial u}{\partial y}
= f'(u)\color{blue}{\left(\frac{x}{z} - \frac{xy}{z^2}q\right)}
\end{align}
$$
This leads to
$$
\color{red}{p} : \color{red}{\left(\frac{y}{z} - \frac{xy}{z^2}p\right)} 
= \color{blue}{q} : \color{blue}{\left(\frac{x}{z} - \frac{xy}{z^2}q\right)}
\implies
\color{red}{p} \color{blue}{\left(\frac{x}{z} - \frac{xy}{z^2}q\right)} 
= 
\color{blue}{q} \color{red}{\left(\frac{y}{z} - \frac{xy}{z^2}p\right)}$$
This can be simplified as
$$px = qy
\quad\iff\quad
x\frac{\partial z}{\partial x} - y\frac{\partial z}{\partial y} = 0
$$
A PDE where $f$ completely disappears. I hope this is clear enough.
A: I try this today.
$p= \dfrac{\partial z}{\partial x}$ and       $q= \dfrac{\partial z}{\partial y} $
So 
$ p= \dfrac{\partial f}{\partial x }\left( \dfrac{yz-pxy}{z^2} \right)  $ and 
$ q= \dfrac{\partial f}{\partial y }\left( \dfrac{xz-pxy}{z^2} \right)  $ 
Is that right? After that, I try many minupulations... All attempts, I needed divided by $\dfrac{1}{\left(\dfrac{\partial f}{\partial x }\right)}$ or divided by $\dfrac{1}{\left(\dfrac{\partial f}{\partial y }\right)}$and this seems wrong to me. 
Anybody can Help?
