Calculation in geometrical interpretation of Lagrange's equation

Question

I was reading the geometrical interpretation of Lagrange's equation and there it was written that if $f(x,y,z)=0$ be the solution of the Lagrange's equation $Pp+Qq=R$, then the direction cosines of the normal to the solution surface at any point are proportional to $p,q,-1$. It comes like this:
$ \frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z}$, i.e $-\frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial z}},-\frac{\frac{\partial f}{\partial y}}{\frac{\partial f}{\partial z}},-1$, i.e $ \frac{\partial z}{\partial x},\frac{\partial z}{\partial y},-1$, i.e $ p,q,-1$
My question is how $-1$ comes in this calculation and how $-$ signs disappear in the $3$rd line from the first $2$ fractions?
There may be something silly that I am missing but I really can't figure it out.

Answer 1

At first each fraction was divided by $-\dfrac{\partial f}{\partial z}$. So it becomes

$-\dfrac{\partial f/\partial x}{\partial f/\partial z},\quad-\dfrac{\partial f/\partial y}{\partial f/\partial z},\quad-1 \tag {1}$

Now here it was assumed that $z=\phi(x,y)$. So what we have is
\begin{align} f(x,y,z)&=0 \implies \dfrac{\partial f}{\partial x}+\dfrac{\partial f}{\partial z}\dfrac{\partial z}{\partial x}=0\\ \implies \dfrac{\partial z}{\partial x}&=-\dfrac{\partial f/\partial x}{\partial f/\partial z} \end{align}
Similarly $\dfrac{\partial z}{\partial y}=-\dfrac{\partial f/\partial y}{\partial f/\partial z}$

Now plug these values in $(1)$ and we obtain
$\dfrac{\partial z}{\partial x},\dfrac{\partial z}{\partial y},-1\implies p,q,-1$