Let $f$ be a vector field in $\mathbb{R}^n$, with $f(0) = 0$. If there is at least one eigenvalue $\lambda$ of $Df(0)$ such that $Re(\lambda) < 0$, then there is $x(t)$ such that $\lim_{t \rightarrow +\infty} x(t) = 0$.
I have an idea about how to show this, but I'm having trouble to be formal. I know that the solutions of $x'(t) = f(x)$ in a neighboorhood of the singularity (in our case, the singularity is 0) are really similar to the solutions of $x' = Df(0)x$, which is linear.
Consider $\overline{x} '(t) = Df(0)x$, $\overline{x}(0) = v$, $v$ such that $Df(0)v = \lambda v$. Then, $\overline{x}(t) = ve^{\lambda t}$ is the solution and $\lim_{t \rightarrow +\infty} x(t) = 0$