Multi-variable Chain Rule in Quasi-linear PDEs In trying to understand the general solution of a first-order quasilinear PDE of the form
$$a(x,y,z) \frac{\partial z}{\partial x}+b(x,y,z) \frac{\partial z}{\partial y}=c(x,y,z) $$

What approach is taken to differentiate $F(u,v)$? When I try to apply the chain rule I am finding 
$$\frac{dF}{dx} = \frac{\partial F}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial F}{\partial v}\frac{\partial v}{\partial x}+\frac{\partial F}{\partial z}\frac{\partial z}{\partial x}$$
which is clearly not in agreement with the book derivation. Also, shouldn't the term $\frac{\partial z}{\partial x} = 0$ as z is not dependent on either x nor y?
 A: So chain rule when applied to $F$ dependent on $u, v$ would yield 
$$dF = \frac{\partial F}{\partial u} du + \frac{\partial F}{\partial v} dv $$
Now we assume $u,v$ are dependent on $x,y,z$ where $z$ is dependent on $x, y$:
$$\frac{du}{dx} = \frac{\partial u}{\partial x} + \frac{\partial u}{\partial z} \frac{\partial z}{\partial x} $$
which yields the almost identical cases for $\frac{du}{dy}, \frac{dv}{dx}, \frac{dv}{dy}$.
So differentiating w.r.t $x$ gives:
$$\frac{dF}{dx} = \frac{\partial F}{\partial u} \frac{du}{dx} + \frac{\partial F}{\partial v} \frac{dv}{dx} $$
$$= \frac{\partial F}{\partial u} \left (\frac{\partial u}{\partial x} + \frac{\partial u}{\partial z} \frac{\partial z}{\partial x}  \right ) + \frac{\partial F}{\partial v} \left (\frac{\partial v}{\partial x} + \frac{\partial v}{\partial z} \frac{\partial z}{\partial x}  \right ) $$
Which is identical to that in your book. 
To answer your follow up, notice that the original equation, you are solving for a $z = f(x,y)$, so $z$ is very much dependent on the variables $x,y$.
