So my question is on the derivation of the implicit differentiation (taken from here).
The general chain rule, from here, it says that if we have a function $z$ of $n$ variables, $x_1, x_2,\ldots,x_n$ and each of these variables are in turn a function of $m$ variables, $t_1, t_2,\ldots, t_m$. Then for any $t_i, i=1, 2, \ldots, m$ we have (1) $$ \frac{\partial z}{\partial t_i}=\frac{\partial z}{\partial x_1}\frac{\partial x_1}{\partial t_i}+\frac{\partial z}{\partial x_2}\frac{\partial x_2}{\partial t_i}+\cdots+\frac{\partial z}{\partial x_n}\frac{\partial x_n}{\partial t_i}$$
My question is how does did the differentiation of $F$ with respect to $x$ come up as it did. How did it come up as (2) $$\frac{\partial F}{\partial x}\frac{\partial x}{\partial x}+\frac{\partial F}{\partial y}\frac{\partial y}{\partial x}+\cdots+\frac{\partial F}{\partial z}\frac{\partial z}{\partial x}=0$$ If I match (2) with (1), then it seems that the left hand side is $\frac{\partial F}{\partial x}$ because $F=z, x=x_1, y=x_2, z=x_3$. As far as the initial condition of $F$ having three variables $(x, y, z)$ that in turn are each supposed to be functions of more variables, can it be assumed that $x$ and $y$ are constants; this would be mean effectively that $x=g(t_1, t_2, t_3)=x$ of some function g and $y=y(t_1, t_2, t_3)=y$. So is the left hand side of (2) $\frac{\partial F}{\partial x}$ or some different notation. What does the differentiation of $F$ with respect to $x$ on the left hand side look like in symbols then?
When god created the world there were three independent quantities $x$, $y$, $z$ changing with time, say. Then a physicist came and said: These three quantities are related by a so-and-so equation $$F(x,y,z)=0\ .\tag{1}$$ At any rate, at some instant $t_0$ of time one indeed had $F(x_0,y_0,z_0)=0$. The equation $(1)$ connecting the three quantities $x$, $y$, $z$ reduces the degrees of freedom by $1$. Therefore we have the feeling that, given two values $x$ and $y$, the value of $z$ should be determined by $(1)$, at least in the neighborhood of the point $P_0=(x_0,y_0,z_0)$. This means that there should be a function $z=\psi(x,y)$ defined in a neighborhood $U$ of $(x_0,y_0)$ and giving $\psi(x_0,y_0)=z_0$, such that $$F\bigl(x,y, \psi(x,y)\bigr)=0\qquad\forall (x,y)\in U\ .\tag{2}$$ Sometimes we can explicitly solve $(1)$ for $z$, and then we have this $\psi$ as a a function of the variables $x$ and $y$ "in finite terms". But in most cases we can't. That's where the implicit function theorem comes in. It says that such a function $z=\psi(x,y)$ defined in the neighborhood of $(x_0,y_0)$ indeed exists, and that this $\psi$ is even differentiable. In particular, we are able to compute the partial derivatives $\psi_x(x_0,y_0)$ and $\psi_y(x_0,y_0)$ from the given data. The chain rule allows to argue as follows: From $(2)$ we obtain by differentiating with respect to $x$ and putting $(x,y):=(x_0,y_0)$ afterwards: $$F_{.1}(x_0,y_0,z_0)\cdot 1+F_{.3}(x_0,y_0,z_0)\psi_x(x_0,y_0)=0\ .$$ This gives $$\psi_x(x_0,y_0)=-{F_{.1}(x_0,y_0,z_0)\over F_{.3}(x_0,y_0,z_0)}\ ,\tag{3}$$ under the "technical assumption" $$F_{.3}(x_0,y_0,z_0)\ne0\ .\tag{4}$$ The formula $(3)$ is usually written in the shorthand version $${\partial z\over\partial x}\biggr|_{(x_0,y_0)}=-{F_x\over F_z}\biggr|_{(x_0,y_0,z_0)}\ .$$ Similarly, one has $${\partial z\over\partial y}\biggr|_{(x_0,y_0)}=-{F_y\over F_z}\biggr|_{(x_0,y_0,z_0)}\ .$$ Note that the assumption $(4)$ is essential for all this to hold.