Suppose we have a system of $n$ autonomous ODEs, which we denote as $$\dot x = f(x), $$ where $x \in \mathbb R^n$ so that for $1 \le i \le n$, we can write each ODE as $$ \dot x_i = f_i(x) = f_i(x_1, \dots, x_n). $$ My book on differential equations claims that if we "use one of the components of $x$ (say $x_1$) as a new independent variable (which requires that $f_1(x) \neq 0$)", then "with the chain rule we obtain $n-1$ equations": $$ \frac{dx_2}{dx_1} = \frac{f_2(x)}{f_1(x)} \\ \vdots \\ \frac{dx_n}{dx_1} = \frac{f_n(x)}{f_1(x)}. $$ I don't seem to understand this step enough to fill in the details, i.e. how this "change of variables" looks and how exactly to apply the chain rule to conclude this. One resource I found on the web, which considered the two-dimensional case, reasons that we can apply the chain rule to $x_2(t)$ in the following way: $$ \dot x_2 = \frac{dx_2}{dx_1} \dot x_1 \Rightarrow \frac{dx_2}{dx_1} = \frac{\dot x_2}{\dot x_1} = \frac{ f_2(x_1, x_2)}{f_1(x_1, x_2)}, $$ but this (again) implies that we can somehow write $x_2(t)$ as a function of $x_1$ only.
In Boyce & DiPrima's chapter on autonomous ODEs, they simply write $$ \frac{dx_2}{dx_1} = \frac{dx_2/dt}{dx_1/dt} = \frac{f_2(x_1,x_2)}{f_1(x_1,x_2)},$$ which makes intuitive sense since it is 'formally multiplying by $dt$', but it's all still a bit vague to me.
Can someone help me fill in the details here? Thanks in advance.
Since you know that $x_1'=f_1\neq 0$ then $x_1'$ is either positive or negative along any solution. In other words, $x_1$ is a (strictly) monotonic function of $t$ and there is a well-defined inverse function $t=t(x_1)$ with the same smoothness. Therefore, you can take $x_1$ to be your new independent variable. W.r.t. this new variable the first equation is trivial; $dx_1/dx_1=1$ and is not needed. To get the rest of them, you use the chain rule as you mention in your question. Observe that this procedure allows to reduce by one the number of equations, but some information is lost on the way, namely the "dynamics" in time. This is how you get the equation of the orbits in the two-dimensional case say, but you do not know how your phase point is moving along them.