= Solution
Write $f(u)=au^2/(b^2+u^2)$, with $b>0$. Besides zero, the <fixed points> satisfy $u^2-au+b^2=0$. If $a^2>4b^2$, they are
$$
\boxed{u_\pm=\frac{a\pm\sqrt{a^2-4b^2}}2}.
$$
The <derivative> is $f'(u)=2ab^2u/(b^2+u^2)^2$. At a positive <fixed point>, $b^2+u^2=au$, so $f'(u)=2b^2/(au)=2(1-u/a)$. Thus zero is attracting, $u_-$ is repelling because $f'(u_-)>1$, and $u_+$ is attracting because $0<f'(u_+)<1$.
Moreover,
$$
f(u)-u=-\frac{u(u-u_-)(u-u_+)}{b^2+u^2}.
$$
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>. \b[$u_-$ is the extinction threshold.] The <cobweb plot> illustrates these alternatives.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2010/ii/paper-2-cobweb.png]
{title=Population cobwebs below and above the unstable extinction threshold}
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.
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