Find the general integral of $ px(z-2y^2)=(z-qy)(z-y^2-2x^3).$ $ p=\frac{\partial z}{\partial x} $ and $ q=\frac{\partial z}{\partial y} $
Find the general integral of the linear PDE $ px(z-2y^2)=(z-qy)(z-y^2-2x^3). $
My attempt to solve this is as follows:
$ p=\frac{\partial z}{\partial x} $ and $ q=\frac{\partial z}{\partial y} $
$$px(z-2y^2)+qy(z-y^2-2x^3)=z(z-y^2-2x^3)$$
\begin{align*}
\text{The Lagrange's auxiliary equation is:} \frac{dx}{x(z-2y^2)}=\frac{dy}{y(z-y^2-2x^3)}=\frac{dz}{z(z-y^2-2x^3)}
\end{align*}
Now consider the 2nd and 3rd ratios, 
\begin{align*}
\frac{dy}{y(z-y^2-2x^3)} & =\frac{dz}{z(z-y^2-2x^3)}\\
\implies \frac{dy}{y} & =\frac{dz}{z}\\
\implies \ln(y) & =\ln(z)+\ln(c_1)\\
\implies \frac{y}{z} & =c_1.
\end{align*}
But I am unable to get the 2nd integral surface. Kindly, help me.
Thanks in advance.

Your calculus is correct.
A first family of characteristic curves comes from $\frac{dx}{x(z-2y^2)}=\frac{dy}{y(z-y^2-2x^3)}$ which solution is $z=\frac{1}{c_1}y=c'_1y$
$$\frac{z}{y}=c'_1$$
A second family of characteristic curves comes from 
$$\frac{dx}{x(c'_1y-2y^2)}=\frac{dy}{y(c'_1y-y^2-2x^3)}$$
The solution of this ODE is :
$y=\frac{c'_1}{2}\pm \sqrt{x^3+\frac{(c'_1)^2}{4}+c_2}$
$\left(y-\frac{c'_1}{2}\right)^2-x^3-\frac{(c'_1)^2}{4}=c_2$
$y^2-c'_1y -x^3=c_2$
$$y^2-z-x^3=c_2$$
The general solution of the PDE expressed on the form of implicit equation is :
$$\Phi\left(\:\left(\frac{z}{y}\right)\:,\:\left(y^2-z-x^3\right)\: \right)=0$$
where $\Phi$ is an arbitrary function of two variables.
Or equivalently :
$$z=y\:F(y^2-z-x^3)$$
where $F$ is an arbitrary function.
$\Phi$ or $F$ has to be determined to fit some boundary condition (not specified in the wording of the question).
