I saw a written process to solve the following question: For $C^1$ function $f:U(\subset\Bbb{R}^n)\to \Bbb{R}$, such that ${\partial f\over \partial x_i}\ne 0$ for all $1\le i\le n$, find ${\partial x_1\over \partial x_2}{\partial x_2\over \partial x_3}...{\partial x_n\over \partial x_1}$.
The solution legitimately claimed that a set of equations is achieved:
$\begin{cases} f(x_1(x_2,...,x_n),x_2,...,x_n)=0 \\[2ex] f(x_1,x_2(x_1,x_3,...,x_n),...,x_n)=0 \\[2ex] \vdots \\[1ex] f(x_1,x_2,...,x_n(x_1,...,x_{n-1}))=0 \\[2ex] \end{cases}$
Where, for instance, $x_1(x_2,...,x_n)$ is a function of $n-1$ variables, locally determining $x_1$ in the original equation, as guaranteed by the theorem.
The next thing done in the solution, is differentiation the $i$th equation by $x_{i+1}$ where $1\le i\le n-1$, and the $n-$th equation by $x_1$. Here's what I didn't understand:
$\begin{cases} {\partial f\over \partial x_1}{\partial x_1\over \partial x_2}+{\partial f\over \partial x_2}=0 \\[2ex] {\partial f\over \partial x_2}{\partial x_2\over \partial x_3}+{\partial f\over \partial x_3}=0 \\[2ex] \vdots \\[1ex] {\partial f\over \partial x_n}{\partial x_n\over \partial x_1}+{\partial f\over \partial x_n}=0 \\[2ex] \end{cases}$
How does one arrive at this? Isn't ${\partial f\over \partial x_1}{\partial x_1\over \partial x_2}$ and ${\partial f\over \partial x_2}$ practically the same? Or at least in the same multiplication, and not addition? Can someone put some light on it?
It is just the chain rule: if $f=f(x_1,x_2)$ and $x_1=x_1(x_2)$, then $$ df = \frac{\partial f}{\partial x_1} \frac{\partial x_1}{\partial x_2} + \frac{\partial f}{\partial x_2}. $$ Indeed, $df = \frac{\partial f}{\partial x_1} dx_1 + \frac{\partial f}{\partial x_2}dx_2$, but $dx_1 = x_1'(x_2)dx_2$.
Please notice that I am carefully avoiding to state precise assumptions and theorems.