How can proceed further Question
The given partial differential equation is $$px+qy+pq=xz$$
My approach
Where $p=\frac{\partial z}{\partial x}$ and $q= \frac{\partial z}{\partial y}$
By using charpit method
$$ \frac{dx}{F_p}=\frac{dy}{F_q}=\frac{dz}{pF_p+qF_q} =\frac{-dp}{F_x+pF_z}=\frac{-dq}{F_y+qF_z} $$
Putting the values


$$\frac{dx}{x+q}=\frac{dy}{xy+p}=\frac{dz}{px+qxy+2pq}=\frac{-dp}{p+qy-z-px}=\frac{-dq}{qx-qx}$$
    From the last fraction of $dq$ we get
    $q=a$, where $a$ is arbitary constant.


Substituting the value of $q$ in given PDE we get $p$ as $$p= \frac{x(z-ay)}{x+a}$$
Further the solution of the differential equation be $dz=pdx+qdy$.
From above we get $$dz= \left[\frac{x(z-ay)}{x+a} \right]dx + ady $$
My question
1) How can i integrate the above formed equation, as i am unable to do so.
2) Is there any chance of converting above given equation into standard clairaut's form ?
Any subtle hint is highly appreciated 
Thankyou
 A: If the PDE is 
$$px+qy+pq=xz\:,$$
the Charpit's equations are :
$$\frac{dx}{x+q}=\frac{dy}{y+p}=\frac{dz}{px+qy+2pq}=\frac{-dp}{p-z-px}=\frac{-dq}{q-qx}$$
This is not what you wrote.
Then $z=ay+be^x(x+a)^{-a}$ is not solution of the above PDE.
$$ $$
If the Charpit's equations are :
$$\frac{dx}{x+q}=\frac{dy}{xy+p}=\frac{dz}{px+qxy+2pq}=\frac{-dp}{p+qy-z-px}=\frac{-dq}{qx-qx}$$
The PDE is :
$$px+qxy+pq=xz$$
This is not the PDE written in your question. 
$$\boxed{z=ay+be^x(x+a)^{-a}\quad\text{is solution of}\quad px+qxy+pq=xz}$$
A: From the last equation formed in question
$$dz = \left[\frac{x(z-ay)}{x+a}\right]dx + ady $$
$$\frac{dz}{z-ay}= \frac{x dx}{x+a} + \frac{a dy}{z-ay} $$
And 
$$\frac{dz-ady}{z-ay} = \frac{x+a}{x+a} + \frac{a}{x+a} $$
 Integrating the  above equation
$$ log_e(z-ay) = x -alog_e(x+a) + log_e{b}$$
Where $log_e{b}$ is arbitrary constant
$$ log_e(z-ay) = x+ log_e(x+a)^{-a} +log_e{b} $$
And the final answer be
$$z = be^{x}(x+a)^{-a} +ay$$
edit :- added the missing term $ay$ in Right hand side.
As suggested by JJacquelin
