I am trying to solve this PDE, using Lagrange's method: $p\sqrt x+q\sqrt y =\sqrt z$
Lagrange's auxiliary equation is: $\frac{dx}{\sqrt x}=\frac {dy}{\sqrt y}= \frac{dz}{\sqrt z}$
I already know the answer to this on which is $f(\sqrt x - \sqrt y, \sqrt y - \sqrt z)$ , but I don't know the complete solution and I have tried to solve it but I can't quite get to the same answer.
$$\sqrt{x} \frac{\partial z}{\partial x}+\sqrt{y} \frac{\partial z}{\partial y} =\sqrt z$$ Your general solution on the form of implicit equation : $\quad f(\sqrt x - \sqrt y, \sqrt y - \sqrt z)=0\quad$ is correct.
An equivalent explicit form is : $$ \sqrt y - \sqrt z =\phi( \sqrt x - \sqrt y)$$ where $\phi$ is any differentiable function.
As a consequence : $$z(x,y)=\bigg(\sqrt y - \phi( \sqrt x - \sqrt y)\bigg)^2$$
IN ADDITION :
System of characteristic ODEs : $\quad\frac{dx}{\sqrt x}=\frac{dy}{\sqrt y}=\frac{dz}{\sqrt z}$
First family of characteristic curves, from $\quad \frac{dx}{\sqrt x}=\frac{dy}{\sqrt y}\quad\to\quad \sqrt x-\sqrt y=c_1$
Second family of characteristic curves, from $\quad \frac{dz}{\sqrt z}=\frac{dy}{\sqrt y}\quad\to\quad \sqrt z-\sqrt y=c_2$
General solution : $\quad\sqrt y - \sqrt z =\phi( \sqrt x - \sqrt y)$
$$z(x,y)=\bigg(\sqrt y - \phi( \sqrt x - \sqrt y)\bigg)^2$$