Solve partial differential equation, $$F=xp^2-ypq+y^2q-y^2z=0,\;\;\;\; p = \frac{\partial z}{\partial x},\;\;\;\;q = \frac{\partial z}{\partial y}$$
My attempt:
$F_x = p^2, \\F_y = -pq+2yq-2yz, \\F_p = 2px-yq,\\ F_q = -yp+y^2, \\F_z=-y^2$
Charpit's Equations:
$$\frac{dx}{yq-2px} = \frac{dy}{y(p-y)} = \frac{dp}{p(p-y^2)} = \frac{dq}{q(2y-p)-yq(p-y)-2yz} $$
Taking $dp$, $dy$ terms: $$(p^2-py^2)dy-(py-y^2)dp=0 \\ p(-pdy+ydp)+y^2(pdy-dp) =0\\d(p/y)+dy-\frac{dp}{p}=0 \\ e^{\frac{p}{y}}e^y=p c_1 \\$$
where $c_1$ is some arbitrary constant
Now how to proceed further?
Before proceeding with the solution, let's express $p$ as an explicit function of $y$: $$ e^{p/y}e^y=pc_1 \implies -\frac{p}{y}e^{-p/y}=\frac{ae^y}{y} \implies p(y)=-yW\!\left(\frac{ae^y}{y}\right), \tag{1} $$ where $a=-1/c_1$ and $W$ is the Lambert $W$ function.
From $p=\frac{\partial z}{\partial x}$ and $(1)$ it follows that $$ z=-xyW\!\left(\frac{ae^y}{y}\right)+g(y)=xp(y)+g(y), \tag{2} $$ hence $q=-\frac{\partial z}{\partial y}=xp'(y)+g'(y)$. To determine $g$, we substitute $q$ and $z$ in the PDE $xp^2-ypq+y^2q-y^2z=0$: $$ xp^2-yp(xp'+g')+y^2(xp'+g')-y^2(xp+g)=0 $$ $$ \implies (p^2-ypp'+y^2p'-y^2p)x-ypg'+y^2g'-y^2g=0. \tag{3} $$ The term multiplying $x$ in Eq. $(3)$ is identically zero, since $p$ satisfies the equation $(y^2-py)p'+p^2-py^2=0$ (cf. OP). We are left, therefore, with the following separable ODE for $g$: $$ (y^2-yp)g'-y^2g=0, \tag{4} $$ whose solution is $$ g(y)=b\exp\left(\int\frac{y^2\,dy}{y^2-yp}\right)=b\exp\left(\int\frac{dy}{1+W\!\left(\frac{ae^y}{y}\right)}\right). \tag{5} $$ Plugging $(5)$ into $(2)$, we finally obtain the following complete integral of the PDE: $$ z(x,y)=-xyW\!\left(\frac{ae^y}{y}\right)+b\exp\left(\int\frac{dy}{1+W\!\left(\frac{ae^y}{y}\right)}\right). \tag{6} $$