 Solution

Write $f(u)=a{u}^2/(b^2+u^2)$, with $b>0$. Besides zero, the fixed points satisfy $u^2-au+b^2=0$. If $a^2>4b^2$, they areThe derivative is $f^{\prime }(u)=2a{b}^{2}u/(b^2+u^2{)}^2$. At a positive fixed point, $b^2+u^2=au$, so $f^{\prime }(u)=2b^2/(au)=2(1-u/a)$. Thus zero is attracting, $u_{-}$ is repelling because $f^{\prime }(u_{-})>1$, and $u_{+}$ is attracting because $0<f^{\prime }(u_{+})<1$.

Moreover,For $0<u_0<u_{-}$ the iterates strictly decrease and stay positive. Their limit is a fixed point below $u_{-}$, hence zero. For $u_{-}<u_0<u_{+}$ the iterates increase towards $u_{+}$; for $u_0>u_{+}$ they decrease towards $u_{+}$. Monotonicity of $f$ prevents crossing the limiting fixed point. $u_{-}$ is the extinction threshold. The cobweb plot illustrates these alternatives.

For $a^2<4b^2$ there are no positive fixed points, and $f(u)<u$ for every $u>0$, so all populations tend to zero. At $a^2=4b^2$, the positive fixed point $a/2$ is semistable: initial populations below it tend to zero, and those above it decrease towards it. This is a saddle-node bifurcation in the discrete dynamics.