Transform the partial differential equation with new independent variables 
Transform the following equation to new independent variables $u = x$, $v = x^2+y^2$
$$ y\frac{\partial z}{\partial x} - x \frac{\partial z}{\partial y} = 0  $$

Notce you don't have to actually solve the pde, the problem is to transform it. The official answer is $ \frac{\partial z}{\partial u} = 0 $
My work:
$$ y\frac{\partial z}{\partial x} = x\frac{\partial z}{\partial y} \longrightarrow \frac{1}{x}\frac{\partial z}{\partial x} = \frac{1}{y}\frac{\partial z}{\partial y}  $$
$$ \frac{\partial z}{\partial y} = \frac{\partial z}{\partial v}\frac{\partial v}{\partial y} \longrightarrow \frac{\partial z}{\partial y} = \frac{\partial z}{\partial v}2y $$
Substituting, you get
$$ \frac{1}{x} \frac{\partial z}{\partial x}  = 2\frac{\partial z}{\partial v}    $$
By the same procedure you can get
$$ \frac{\partial z}{\partial x} = \frac{\partial z}{\partial u} $$
Finally, I got

$$ \frac{1}{u}\frac{\partial z}{\partial u} = 2\frac{\partial z}{\partial v} $$

Which is obviously not the official answer. Am I doing something wrong and if not, how do I finish the problem?
Thanks.
 A: In your procedure the  step $\frac{\partial z}{\partial x} = \frac{\partial z}{\partial u}$
does not true.

By chain rule:
$$\frac{\partial z}{\partial x} = \frac{\partial z}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial z}{\partial v}\frac{\partial v}{\partial x}$$
$$\frac{\partial z}{\partial y} = \frac{\partial z}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial z}{\partial v}\frac{\partial v}{\partial y}$$

$\implies$ $$\frac{\partial z}{\partial x} =\frac{\partial z}{\partial u}(1)+\frac{\partial z}{\partial v}(2x)=\frac{\partial z}{\partial u}+2x\frac{\partial z}{\partial v}$$ and
$$ \frac{\partial z}{\partial y} =\frac{\partial z}{\partial u}(0)+\frac{\partial z}{\partial v}(2y)=2y\frac{\partial z}{\partial v}$$
So $y\frac{\partial z}{\partial x} - x \frac{\partial z}{\partial y} = 0\implies  y\frac{\partial z}{\partial u}+2yx\frac{\partial z}{\partial v}-2xy\frac{\partial z}{\partial v}=0\implies \frac{\partial z}{\partial u}=0$
A: $\frac{\partial z}{\partial x} = \frac{\partial z}{\partial u}\frac{d u}{d x}+\frac{\partial z}{\partial v}\frac{\partial v}{\partial x}$

$$\frac{\partial z}{\partial x} = \frac{\partial z}{\partial u}\cdot1 + \frac{\partial z}{\partial v}\cdot2x $$

Similarly,

$$\frac{\partial z}{\partial y} = 0 + \frac{\partial z}{\partial v}\cdot2y$$

Now, $y\frac{\partial z}{\partial x} - x\frac{\partial z}{\partial y} = 0$
$y\frac{\partial z}{\partial u} + 2xy\cdot\frac{\partial z}{\partial v} -2xy\cdot \frac{\partial z}{\partial v}=0\implies y\frac{\partial z}{\partial u}=0$
As, $y$ need not be necessarily $0$,

$$\frac{\partial z}{\partial u}=0$$

A: $${\partial z\over\partial x}={du\over dx}{\partial z\over\partial u}+{dv\over dx}{\partial z\over\partial v}=(1){\partial z\over\partial u}+(2x){\partial z\over\partial v}$$
$${\partial z\over\partial y}={du\over dy}{\partial z\over\partial u}+{dv\over dy}{\partial z\over\partial v}=(0){\partial z\over\partial u}+(2y){\partial z\over\partial v}$$
$$0=y{\partial z\over\partial x}-x{\partial z\over\partial y}=y\left({\partial z\over\partial u}+2x{\partial z\over\partial v}\right)-x\left(2y{\partial z\over\partial v}\right)=y{\partial z\over\partial u}$$
Hence ${\partial z\over\partial u}=0$
The more important point is that I don't think you understand how to use the chain rule in multivariable calculus - perhaps you should check how it works
