I have two differential equations:
\begin{equation} x' = x(1-y)\\ y' = y(x-1) \end{equation}
After some tricks, where you can use:
\begin{equation} \frac{dy}{dx} = \frac{y(x-1)}{x(1-y)} \rightarrow \int \frac{(1 - y)}{y} dy = \int \frac{(x-1)}{x}dx. \end{equation}
This eventually leads to: \begin{equation} - f(y, c) = \log(y) - y + C = x - \log(x) + C = f(x, y) \end{equation}
How can I determine for which values of $C$ the functions solutions?
From
$$ \ln y-y=x-\ln x+C $$
Assuming $x > 0, y > 0$ we have
$$ ye^{-y}= e^C \frac{e^x}{x} $$
or
$$ e^C = x y e^{-(x+y)} $$
now $x y e^{-(x+y)}$ has a maximum at $x = y = 1$ then
$$ 0 < e^C \le \frac{1}{e^2}\Rightarrow -\infty < C\le -2 $$