= Solution
The first <equilibrium> equation gives $y=x$, and the third gives $z=x^2/b$. The second then becomes $x(r-1-x^2/b)=0$. Hence the <equilibria> are
$$
\boxed{O=(0,0,0),\qquad C_\pm=(\pm\sqrt{b(r-1)},\pm\sqrt{b(r-1)},r-1)\quad(r>1).}
$$
At the origin, one <eigenvalue> is $-b$ and the remaining two solve $\lambda^2+(\sigma+1)\lambda+\sigma(1-r)=0$. Their discriminant is $(\sigma-1)^2+4\sigma r>0$. For $r<1$ both are negative, so the origin is <linearly stable>, with all <eigenvalues> real and negative. For $r>1$ one is positive and one negative, so it is a <saddle equilibrium> with one unstable and two stable directions. At $r=1$ its <eigenvalues> are $0,-(\sigma+1),-b$: <linearization> has a neutral direction and cannot alone prove nonlinear stability. Given the source's global attraction for $r<1$, there are no nontrivial <limit cycles> or persistent chaotic invariant dynamics in that regime; all initial states tend to the origin.
To establish the local pitchfork, one needs the simple <eigenvalue> to cross zero, the two symmetry-related branches to exist and acquire the expected local stability, and nondegeneracy of the nonlinear coefficient. The <derivative> of the crossing <eigenvalue> is $\sigma/(\sigma+1)>0$. At $C_\pm$, the <characteristic polynomial> is
$$
\boxed{\lambda^3+(\sigma+b+1)\lambda^2+b(\sigma+r)\lambda+2\sigma b(r-1)=0.}
$$
For $r>1$ sufficiently close to one, all coefficients are positive and $(\sigma+b+1)b(\sigma+r)>2\sigma b(r-1)$, so the cubic <Routh-Hurwitz criterion> makes both new <equilibria> stable. Together with the <symmetry> $(x,y,z)\mapsto(-x,-y,z)$ and the square-root branches, this yields the supercritical pitchfork. A local centre-manifold coordinate proportional to $x$ has leading <normal form> $\dot q=\frac\sigma{\sigma+1}(r-1)q-\frac\sigma{b(\sigma+1)}q^3+\cdots$, explicitly displaying the nonzero stabilizing cubic coefficient. At $r=1$ that cubic also gives weak local nonlinear attraction along the centre direction.
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