How to differentiate a given unknown function involving three variables and two arbitrary constants? I was reading 

Elements of partial differential equations by IAN N. SNEDDON

and found out the following:
It is trying to generalise the function given which leads to a pde of first order. It says to consider the function
$$F(x,y,z,a,b) = 0$$
where $a$ and $b$ are constants, $x$ and $y$ are the independent variables and $z$ is the dependent variable.
Then it differentiated this equation w.r.t. $x$ and $y$ to obtain two more equations:
$$ \frac{\partial F}{\partial x} + p \frac{\partial F}{\partial z} = 0 $$
$$ \frac{\partial F}{\partial y} + q\frac{\partial F}{\partial z} = 0 $$
How can this happen? Can someone please explain? How can you differentiate an unknown function as $F$. 
Here $p$ and $q$ are the partial derivatives of $z$ w.r.t $x$ and $y$ respectively.
 A: Suppose we have the function
$$ F(u,v,z) = 0 $$
where $u = u(x,y)$, $v = v(x,y)$, $z = z(x,y)$ are all functions of $x$ and $y$. The multi-variable chain rule states
$$ \frac{\partial F}{\partial x} = \frac{\partial F}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial F}{\partial v}\frac{\partial v}{\partial x} + \frac{\partial F}{\partial z}\frac{\partial z}{\partial x} = 0 $$
$$ \frac{\partial F}{\partial y} = \frac{\partial F}{\partial u}\frac{\partial u}{\partial y} + \frac{\partial F}{\partial v}\frac{\partial v}{\partial y} + \frac{\partial F}{\partial z}\frac{\partial z}{\partial y} = 0 $$
Now let $u(x,y) = x$, and $v(x,y) = y$, then
$$ \frac{\partial u}{\partial x} = 1, \quad \frac{\partial u}{\partial y} = 0 $$
$$ \frac{\partial v}{\partial x}
 = 0, \quad \frac{\partial v}{\partial y} = 1 $$
Now the equations reduce down to
$$ \frac{\partial F}{\partial x} = \frac{\partial F}{\partial u} + \frac{\partial F}{\partial z}\frac{\partial z}{\partial x} = 0 $$
$$ \frac{\partial F}{\partial y} = \frac{\partial F}{\partial v} + \frac{\partial F}{\partial z}\frac{\partial z}{\partial y} = 0 $$
