I have to solve this problem:
Let $D\subset \mathbb{R}^n$ compact and consider the map $$ \dot{x}_i = \prod_{i\neq j}^n x_j, \quad i =1,...n $$ defining the flow of a dynamical system. We use a coordinate transform $$ \Phi_i: (x_i) \mapsto \frac{1}{x_i} \prod_{i\neq j}^n x_j =: y_i $$ Prove that the system of ODEs is transformed into the quadratic system $$ \dot{y} = y_i(-2y_i + \sum_{i=1}^n y_j), i = 1,...,n $$ I dont know how I can prove that. I am not even sure what the question really means. The way I would make use of it, would be to solve the quadratic ODE for y and substitute back into the first ODE, like so: assume $y$ to be the solutionvector with its componentfunctions to be the solutions to the $i$-th quadratic ODE. Then we can write $$ y_i = \frac{\dot{x_i}}{x_i} \Rightarrow \dot{x_i} = x_iy_i $$ which we maybe could solve with separation of variables. But this does not prove, what was asked for. I made no connection to the initially given problem.
Taking the first relation that
$$ y_i \;\; =\;\; \frac{\dot{x}_i}{x_i}, $$
we can rearrange the rest of the equation to eliminate all the various instances of $x$ in the equation. It is a straightforward but tedious computation to show that
$$ \frac{d}{dt}(x_iy_i)\;\; =\;\; \dot{x}_iy_i + x_i \dot{y}_i \;\; =\;\; \sum_{k=1 \\ k\neq i}^n \left (\dot{x}_k \prod_{j\neq i,k}^n x_j \right ). $$
Now noticing that $\dot{x}_i = x_iy_i$ our equation then becomes
$$ x_i\dot{y}_i + x_iy_i^2 \;\; =\;\; \sum_{k=1 \\ k\neq i}^n \left (\dot{x}_k \prod_{j\neq i,k}^n x_j \right ) $$
Isolating $\dot{y}_i$ we find \begin{eqnarray*} \dot{y}_i & = & - y_i^2 + \sum_{k=1 \\ k\neq i}^n \left (\dot{x}_k \frac{1}{x_i}\prod_{j\neq i,k}^n x_j \right ) \\ & = & - y_i^2 + \sum_{k=1 \\ k\neq i}^n \left (\dot{x}_k \frac{1}{x_ix_k}\prod_{j\neq i}^n x_j \right ) \\ & = & -y_i^2 + \sum_{k=1 \\ k\neq i}^n \frac{\dot{x}_k}{x_k}y_i \\ & = & y_i \left (-y_i + \sum_{k=1 \\ k\neq i}^n y_k \right ) \\ & = & y_i \left (-2y_i + \sum_{k=1}^n y_k \right ). \end{eqnarray*}