<s>[INST] Solve the PDE $(xz-y)p+(yz-x)q=xy-z$ using lagrange method I have the following PDE,
$$(xz-y)p+(yz-x)q=xy-z$$where $p=z_x,\quad q=z_y$
Now having a hard time to get two solution from,
$$\frac{dx}{xz-y}=\frac{dy}{yz-x}=\frac{dz}{xy-z}$$
I can't think of any multipliers trick which can help me to get one. In fact, $dx-dy$ or $dx+dy$ also couldn't help here. I guess the $xy-z$ term is the main culprit.
Any solution or hint will be appreciated. It will be a great help if you also describe how to guess the trick (if you used so).
Thanks in advance.
update
In fact, I got two more problems where I faced the similar issue:
$$\left(x z+y^{2}\right) p+\left(y z-2 x^{2}\right) z=-\left(2 x y+z^{2}\right)\qquad (1)$$
gives
$$\frac{d x}{x z+y^2}=\frac{d y}{y z-2 x^{2}}=\frac{d z}{-\left(2 x y+z^{2}\right)}$$

$$(y+3 z)p+(z+5 x)q=x+7 y\qquad (2)$$
$$
\begin{align}
&\frac{d x}{y+3 z}=\frac{d y}{z+5 x}=\frac{d z}{x+7 y}\\
\implies&\frac{d x-3d y}{y-15x}=\frac{d y-5d z}{z-7 y}=\frac{d z-7d x}{x-21 z}
\end{align}
$$
 [/INST] This is not a complete answer rather expanding the @ShengtongZhang's one as the OP commented there,
characteristic equation of that PDE,
$$
\begin{align}
&\frac{dx}{xz-y}=\frac{dy}{yz-x}=\frac{dz}{xy-z}=dt\\
&x'=xz-y,\quad y'=yz-x, \quad z'=xy-z
\end{align}
$$
Now, inspecting the $x'y-y'x$ term,
$$
\begin{align}
x'y-y'x&=(xz-y)y-(yz-x)x\\
&=xyz-y^2-xyz+x^2\\
&=x^2-y^2 \text{(free of $z$ variable)}
\end{align}
$$
Which give some idea to guess one solution as $\frac{x}{y}$,
$$
\begin{align}
\frac{d}{dt}\left(\frac{x}{y}\right)&=\frac{x'y-y'x}{y^2}\\
&=\frac{x^2-y^2}{y^2} \text{(solvable)}
\end{align}
$$
</s>