$\displaystyle f(xy,z-2x)=0$ satisfies $x \dfrac{\partial{z}}{\partial{x}}-y \dfrac{\partial{z}}{\partial{y}}=2x$ Show that $\displaystyle f(xy,z-2x)=0$ satisfies under certain conditions, the equation $x \dfrac{\partial{z}}{\partial{x}}-y \dfrac{\partial{z}}{\partial{y}}=2x$. What are these conditions?
Attempt: $\beta=z-2x$
$x \dfrac{\partial{\beta}}{\partial{x}}=x \dfrac{\partial{z}}{\partial{x}}-2x$
$y \dfrac{\partial{\beta}}{\partial{y}}=y \dfrac{\partial{z}}{\partial{y}}$
using the given equation we get
$x \dfrac{\partial{\beta}}{\partial{x}}=y \dfrac{\partial{\beta}}{\partial{y}}$
 A: Finding implicitly from $f(xy, z-2x) = 0 $ the derivative $$\frac {\partial z} {\partial x}= -\frac {D_1f(xy, z-2x)y-2D_2f(xy, z-2x)}{D_2f(xy, z-2x)}$$ and the derivative $$ \frac {\partial z} {\partial y }= -\frac {D_1f(xy, z-2x)x} {D_2f(xy, z-2x)} ,$$ we have
 $$x \dfrac{\partial{z}}{\partial{x}}-y \dfrac{\partial{z}}{\partial{y}}= 2x. $$
A: Let $u=xy,v=z-2x$, then, $\frac{\partial u }{\partial x}=y$, $\frac{\partial u }{\partial y}=x$ $\frac{\partial u }{\partial z}=0$ 
Similarly, $\frac{\partial v}{\partial x}=\frac{\partial z}{\partial x}-2$, $\frac{\partial v}{\partial y}=\frac{\partial z}{\partial y}$ ,$\frac{\partial v}{\partial z}=1$
Now, $\frac{\partial f}{\partial x}=\frac{\partial f}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial x}=\frac{\partial f}{\partial u}y+\frac{\partial f}{\partial v}(\frac{\partial z}{\partial x}-2)=0$
$\frac{\partial f}{\partial y}=\frac{\partial f}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial y}=\frac{\partial f}{\partial u}x+\frac{\partial f}{\partial v}\frac{\partial z}{\partial y}=0$
From the second equation, we have $\frac{\partial f}{\partial u}=-(\frac{1}{x}\frac{\partial z}{\partial y})\frac{\partial f}{\partial v}$
Substitute this into the first equation and obtain;
$-(\frac{y}{x}\frac{\partial z}{\partial y})\frac{\partial f}{\partial v}+\frac{\partial f}{\partial v}(\frac{\partial z}{\partial x}-2)=(-\frac{y}{x}\frac{\partial z}{\partial y}+\frac{\partial z}{\partial x}-2)(\frac{\partial f}{\partial v})=0$
Observe that $-\frac{y}{x}\frac{\partial z}{\partial y}+\frac{\partial z}{\partial x}-2=0$ gives the result after multiplication by $x$.
Also, note that we did not use $\frac{\partial f}{\partial z}$ because it did not produce any result as you can check.
