Consider the potential $$ V(x,y,z)=x^4-x^2+xy+y^2+yz+z^2. $$ and the gradient system $$ \dot{\begin{pmatrix}x\\y\\z\end{pmatrix}}=-\text{grad}V(x,y,z). $$ I already determined the equilibria: The equilibria are $$ O_1=(0,0,0),\qquad O_2=(3z_0,-2z_0,z_0),\qquad (-3z_0,2z_0,-z_0), $$ where $z_0^2=2/27$.
Now, the aim is to say which kind of stability the equilibria have.
Linearizing in $O_1$, one sees that the characteristic equation has two negative and one positive root, i.e. $O_1$ is a saddle.
Next, one sees that the system is symmatric with respect to $(x,y,z)\to (-x,-y,-z)$. Since the points $O_2$ and $O_3$ are symmetric, they have the same stability type.
Now, there is one conclusion I cannot understand:
The potential must have at least one minimum which corresponds to a stable equilibrium. $O_1$ is not stable, so both $O_2$ and $O_3$ are stable.
Why does the potential $V(x,y,z)$ have at least one minimum and why does this correspond to a stable equilibrium of the system?
The potential goes to $\infty$ as $x,y,z \to \infty$. Therefore, on a sufficiently large ball, $V$ must attain its minimum somewhere in the interior. This must be a critical point and it cannot be $O_1$. Hence this must happen at $O_2$ or $O_3$, and due to the symmetry, it must happen at both.
Now at a minimum, the Hessean is positive semidefinite, and thus the linearization of your system has all non-positive eigenvalues, implying linear stability (almost). If you compute the linearization, you will see that the eigenvalues are actually all negtaive at $O_2$ and $O_3$, implying nonlinear asymptotic stability.