How to solve linear partial differential equation?

Given that $P, Q, R$ is a function of $x, y, z$, we have $$P {\partial z \over \partial x} + Q{\partial z \over \partial y} = R \hspace{2 cm} (1)$$ The book says that $u = \phi(v)$ is the solution of $(1)$ if $\phi(u, v) = 0$ is the solution of $${dx \over P} = {dy \over Q} ={dz \over R} \hspace{2 cm} (2)$$ where $\phi$ is an arbitrary function. I don't understand why my book is giving the solution as $\phi(u, v) = 0$ sometimes while $u = \phi(v)$ at other times.

Examples: Let $p = {\partial z \over \partial x}$ and $q = {\partial z \over \partial y}$

Problem 1:- ${y^2zp \over x} + xzq = y^2$, Solution: $f(x^2-z^2, x^3-y^3) = 0$
Problem 2:- $(x+2z)p + (4zx-y)q = 2x^2+y$, Solution: $x^2-y-z =f(xy-z^2)$

When do we have a solution of type $\phi(u, v) = 0$ and when is it of type $u = \phi(v)$?

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It's worth recalling that the vector $(-z_x,-z_y,1)$ is a normal vector to the surface $z=z(x,y)$. Our equation (1) says precisely that the normal vector must be orthogonal to the given vector field $(P,Q,R)$. This makes it natural to look for solutions in the implicit form $F(x,y,z)=0$, because the orthogonality relation becomes $\nabla F\perp (P,Q,R)$. There is a lot of freedom here: the set of possible gradients $\nabla F$ is two-dimensional at every point. We can hope to cover all solutions by choosing two "independent" ones, $u(x,y,z)=0$ and $v(x,y,z)=0$, and then letting $F=\phi(u,v)$ where $\phi$ can be any smooth function.
One practical way to find $u$ and $v$ is to look at the fields such as $(-Q,P,0)$, $(-R,0,P)$ and $(0,-R,Q)$ (which are orthogonal to $(P,Q,R)$ by design) and try to make them conservative by introducing a scalar factor (aka integrating factor). Let's try this in problem 1. We have the field $(y^2z,x^2z,xy^2)$ (no harm in multiplying by $z$ throughout). Noticing the common factor of $y^2$, we go for $(-R,0,P)=y^2(-x,0,z)$ where $(-x,0,z)=\nabla (-x^2+z^2)/2$. This gives $u$. Similarly, there is a common factor of $z$ in the first two components, and without it we get $(-x^2,y^2,0)$, which gives $v$.
But in problem 2 this won't work because we can't get rid of any of three variables by isolating it to a scalar factor. Instead we can try additive cancellation. For example, adding $Q+R$ cancels $y$, and the result is nothing but $2xP$. So, $(-2x,1,1)\perp (P,Q,R)$ which gives $u=-x^2+y+z$ as one of solutions. Similarly, $yP+xQ$ gets rid of some terms and luckily for us, $yP+xQ=2zy+4zx^2=2zR$. So, $(y,x,-2z)\perp (P,Q,R)$, and we get $v=xy-z^2$. Just as in problem 1, the equation $\phi(-x^2+y+z,xy-z^2)=0$ represents an (implicit) solution for any smooth function $\phi$.
Conclusion: Problem 2 should have solution of the form $\phi(-x^2+y+z,xy-z^2)=0$, just as problem 1. Of course, this includes $\phi(u,v)=u+f(v)$ as a special case. It's possible that in your textbook only this special case was needed for some reason.