Let $f: \mathbb{R}^n \rightarrow \mathbb{R}^n$ be continuously differentiable, let $\dot x = f(x)$ be an autonomous ODE, and let $x_* \in \mathbb{R}^n$ be a critical point.
Question: I want to show or find a counterexample to the claim that if $(Df)(x_*)$ has an eigenvalue $\lambda$ with $\text{Re}(\lambda) > 0$, then the point $x_*$ is not stable.
For $n = 1$ this is true, but I am not sure about higher dimensions.
My motivation for this question is the following: If $\dot x = f(x) = Ax$ was a linear system with $A \in \mathbb{R}^{n \times n}$ then the claim were true. For real eigenvalues this can be seen as follows: Bring $A$ into Jordan normal form, $A = V J V^{-1}$. The solutions to the linear ODE are of the form $e^{At} x_0 = V e^{tJ} V^{-1} x_0$. Then pick some eigenvector $v$ to an eigenvalue with strictly positive real part as the initial value. Then we have
$$ e^{At} v = V e^{tJ} V^{-1} v = V (0, \ldots, 0, e^{\lambda t}, 0, \ldots, 0)^T = e^{t \lambda} v$$
Since $\lambda > 0$, we have $\Vert x(t) \Vert \rightarrow \infty$ for $t \rightarrow \infty$. Since we may choose the start value, which was an eigenvector to $\lambda$ with arbitrarily small norm, this shows instability.
For complex eigenvalues this ought to be not much different, except for some rotational terms.
We know that if all the eigenvalues have strictly negative real part, then $x_*$ is stable (even exponentially, asymptotically stable). This uses the same result for the linear case, and then applies this to the Taylor expansion of $f$. However, for this to work, one needs a modified version that takes into account the higher order Taylor rest. This is done using Gronwoll's Lemma. Also, if the largest real part of the eigenvalues is $0$, anything can happen; even in dimension $1$.