This is an excerpt from a dynamical systems paper:
They provide a proof of this Lemma, and numerical simulations also show it should be true. It's clear the equilibrium point on each axis is $\left(\sqrt{\frac{l}{-a}},0,0\right)$, $\left(0, \sqrt{\frac{l}{-a}},0\right)$, and $\left(0,0,\sqrt{\frac{l}{-a}}\right)$. I was doing linear stability analysis, and I started by computing the Jacobian:
$$ J \equiv \left(\begin{array}{rr} \,{l+3ax^2+by^2+cz^2} & 2bxy & 2cxz \\ 2cxy & \,l+3ay^2+bz^2+cx^2 & 2byz \\ \, 2bxz & 2cyz & {l+3az^2+bx^2+cy^2} \end{array}\right) $$
Evaluating the matrix at the first fixed point for example gives:
$$ \left(\begin{array}{rr} \,{-2l} & 0 & 0 \\ 0 & \,-2l & 0 \\ \, 0 & 0 & {-2l} \end{array}\right) $$
For $l=1$, this says that all the eigenvalues are negative, so that this fixed point is stable from all directions (and in fact is independent of what $a,b,c$ are. But the paper provides a proof that a cycle should exist, so I'd expect each point to be a saddle. I must've made a mistake in the Jacobian, but for the life of me, I haven't been able to see where it is.
Look at the middle entry in the Jacobian: At the first fixed point it is not $-2\ell$, it is$$\ell \left( 1-\frac{c}{a} \right)$$ not $-2\ell$. Similarly the bottom right is $$\ell \left( 1-\frac{b}{a} \right)$$. The latter is not negative, because $\frac ba < 1$, so this point is not a stable point, it is a saddle.