 Solution

Write $f(N)=N[r/(1+b{N}^2)-\delta ]$. A steady state is a nonnegative zero of $f$. If $r<\delta$, the bracket is strictly negative for every $N\ge 0$, so only $N=0$ exists. If $r>\delta$, there are exactly two nonnegative steady states,For an autonomous scalar ordinary differential equation, a steady state with $f^{\prime }<0$ is locally asymptotically stable, while one with $f^{\prime }>0$ is unstable. HereAt the positive steady state, use $1+b(N^{\ast }{)}^2=r/\delta$ to obtainTherefore zero is stable for $r<\delta$ and unstable for $r>\delta$, whereas the positive steady state is stable whenever it exists. At the omitted equality $r=\delta$, $f(N)=-\delta b{N}^3/(1+b{N}^2)$; zero is nonhyperbolic but still attracts every nonnegative initial state.