Suppose I am studying the stability of a nonlinear system using the Lyapunov inverse theorem, and suppose I get a jacobian matrix that is singular.
From the theory, I know that this implies that the equilibrium is not isolated.
Can I conclude something about the stability from this fact?
On
No, it does not imply that the equilibrium is not isolated. For example, consider
$$ \eqalign{ \dot{x} &= x\cr \dot{y} &= y^3\cr} $$
But as for your question, if an equilibrium is not isolated, it can't be asymptotically stable. This is obvious from the definition of asymptotically stable: any neighbourhood of your equilibrium contains points, namely other equilibria, that are not attracted to your equilibrium.
The standard textbook example for this question is
$$ \dot{x} = -x^3 $$
which is asymptotically stable but has singular Jacobian at $x=0$ and
$$ \dot{x} = x^3 $$
which is unstable but has also a singular Jacobian at $x=0$. So if the Jacobian is singular the only thing you can conclude about stability is that the Jacobian test is inconclusive.
Note that both systems have an isolated equilibrium even though the Jacobian is singular.