 Solution

The only positive equilibrium solves $1-y=0$ and $x-1=0$, hence isAway from a nullcline,This is separable:Integration gives the first integralThus one convenient conserved quantity is

Its gradient isso $(1,1)$ is its only stationary point in the positive quadrant. Its Hessian matrix iswhich is positive definite everywhere. Hence $(1,1)$ is a strict, indeed global, minimum.

The nearby level curves of $E$ are closed curves surrounding this minimum. Since $E$ is constant along every trajectory, a solution starting on a sufficiently small nearby level set cannot leave the region bounded by a slightly larger level set. This proves Lyapunov stability of the equilibrium: solutions initially close to $(1,1)$ remain close for all time.