The question is about the system of equations
$$\frac{dx}{dt} = y - 1, \frac{dy}{dt} = -xy$$
and I'm trying to make a phase picture by finding the orbits. I'm not sure if orbit is the correct term, but I mean the following: if $\gamma: t \mapsto (x(t), y(t))$ is a solution of the system with initial values $x(t_0) = x_0, y(t_0) = y_0$ for $t \in I$ for some interval $I$, then the orbit is $\gamma[I]$. The solution is given locally by solving $\frac{dy}{dx} = \frac{-xy}{y - 1}$, which gives $y \ln|y| = y_0 \exp(y_0 + \frac{1}{2}(x_0^2 - x^2))$. I just don't see how we can extract the orbit from this.
The stationary points are $(0, 1)$ because that is the solution of $y - 1 = 0$ and $-xy = 0$.