The entire problem reads:
Consider the predator-prey model $\dot x = \left( 4-x-\frac{2y}{1+x} \right)$, $\dot y = y(x-1)$. Assume that all positive solutions are bounded. (a) Find all critical points and determine their local stability. (b) Show that this system has a limit cycle in the first quadrant.
I have already found all of the equilibrium points, the important of which is an unstable focus at $(1,3)$. I attempted to show part (b) using some phase plane analysis, but I have hit a wall and I'm not sure where to go next. I started by plotting the parabola $y = \frac{(4-x)(1+x)}{2}$ because that's where $\dot x = 0$, but it didn't help me gain much. Any and all advice on how to do the rest of part (b) would be greatly appreciated!
Let me repeat, with suitable modifications, my answer to the OP's earlier question.
Assume that all solutions starting in $\mathbb{R}^2_{+} := \{\, (x, y) : x \ge 0,\ y \ge 0 \,\}$ are bounded for $t > 0$ (this would require a separate proof). It follows then that for any such solution, its domain contains $[0,\infty)$ and its $\omega$-limit set is compact and nonempty.
Let $L$ stand for the $ω$-limit set of some point, $(x_0,y_0)$, sufficiently close to the unstable focus $(1,3)$. By the Poincaré–Bendixson theorem, as there are finitely many equilibria, $L$ is either a limit cycle, or a heteroclinic cycle, (EDIT: or a homoclinic loop), or an equilibrium.
There are two equilibria, $(4,0)$ and $(1,3)$. The first of them is an unstable focus, so it cannot belong to any heteroclinic cycle (because a heteroclinic cycle (EDIT: or a homoclinic loop) must contain an equilibrium that is an $\omega$-limit point for some other point). Consequently, there are no heteroclinic cycles (EDIT: or homoclinic loops) at all.
So, $L$ is either a periodic orbit, or equals $\{(4,0)\}$. We proceed now to excluding the latter. The linearization of the vector field at $(4,0)$ has matrix $$ \begin{bmatrix} -1 & -\frac25 \\ 0 & 3 \end{bmatrix}. $$ There are two real eigenvalues, $-1$ and $3$, of opposite signs, so $(4,0)$ is a hyperbolic saddle. Its stable manifold is tangent to an eigenvector corresponding to $-1$, that is, to $(1,0)$. I claim that the stable manifold is the $x$-axis, minus $(4,0)$. Indeed, on the $x$-axis we have $\dot{x} = 4 - x$, $\dot{y} \equiv 0$, so for any $(x_1,0)$ we have $\omega((x_1,0)) = \{(4,0)\}$. Now, for a hyperbolic saddle its stable manifold is just the set of those points whose (unique) $\omega$-limit point is the saddle. Hence, if $L = \{(4,0)\}$ then the positive semitrajectory of $(x_0, y_0)$ must belong to the $x$-axis, which contradicts the uniqueness of the initial value problem (notice that $y_0 > 0$).
We have thus shown that $L$ is a periodic orbit.