Past exam of the algebra 1 and differential equations 1 course of the University of Oxford / 2015 / 4 / b / ii

Intersect the two nullcline families. The axes yield $(0,0)$ and $(1,0)$, and $y=1$ with $xy=2$ yields $(2,1)$. The possible remaining intersection would have $x+y=1$ and $xy=2$, requiring $x^2-x+2=0$, whose discriminant is $-7$. Hence there are no further real equilibria.

The Jacobian matrix isEvaluate it at each equilibrium:

All three are hyperbolic equilibria, so their classifications persist for the nonlinear system. Immediately to the right of $(2,1)$ on $y=1$, the velocity is vertically downward, showing that the attracting spiral turns clockwise.