Find complete integral of $(y-x)(qy-px) = (p-q)^{2}$ 
Find complete integral for partial differential equation $(y-x)(qy-px) = (p-q)^{2}$ where  $p={ \partial  z \over \partial x},q={ \partial  z \over \partial y}$.

My attempt:
The given equation is  f(x,y,z,p,q) = $(y-x)(qy-px) - (p-q)^{2}$
The Charpit's auxilary equation will be given by 
$${dp \over 2px-(p+q)y}={dq \over 2qy-(p+q)x}={dx \over {2(p-q)-x2+xy}}={dy \over {-2(p-q)+xy-y2}}$$ 
From here I tried to proceed but no solvable fraction turned out.

Hint:
Let $\begin{cases}u=x+y\\v=x-y\end{cases}$ ,
Then $\dfrac{\partial z}{\partial x}=\dfrac{\partial z}{\partial u}\dfrac{\partial u}{\partial x}+\dfrac{\partial z}{\partial v}\dfrac{\partial v}{\partial x}=\dfrac{\partial z}{\partial u}+\dfrac{\partial z}{\partial v}$
$\dfrac{\partial z}{\partial y}=\dfrac{\partial z}{\partial u}\dfrac{\partial u}{\partial y}+\dfrac{\partial z}{\partial v}\dfrac{\partial v}{\partial y}=\dfrac{\partial z}{\partial u}-\dfrac{\partial z}{\partial v}$
$\therefore v\left(v\dfrac{\partial z}{\partial u}+u\dfrac{\partial z}{\partial v}\right)=4\left(\dfrac{\partial z}{\partial v}\right)^2$
