Let $U$, $V$ be open sets in $\mathbb{R}^m$, and $\phi : U \rightarrow V$ be a diffeomorphism. Denote the inverse map by $\psi$. Let $f: U \rightarrow \mathbb{R}^m$ smooth.
Define a map $g:V \rightarrow \mathbb{R}^m$ as $g(x) := D \phi (\psi (x)) f(\psi(x))$, where $D\phi$ denotes the Jacobian matrix.
Suppose that $f(x_0)=0$ for some $x_0 \in U$, so that $g(\phi(x_0))=0$ by definition.
Then the following holds.
All the eigenvalues of $Df (x_0)$ has negative real parts if and only if the same condition holds for $Dg(\phi(x_0))$.
Proof: It is well known that $x_0$ is an exponentially stable equilibrium of the ODE $\dot x = f(x)$ if and only if $Df (x_0)$ has eigenvalues with negative real parts. The claim follows immediately from this equivalence.
Question: Is there a proof of the assertion without using any ODE theory?
Yes. Write down the product rule and chain rule and use two facts: $f(x_0)=0$ gets rid of the horrible term; and $D\phi(\psi(x)) = (D\psi(x))^{-1}$. Then you get the fact that $Dg(\phi(x_0)) = P^{-1}Df(x_0)P$ for $P=D\psi(x_0)$. Thus, the two matrices are similar (i.e., represent the same linear map) and, in particular, have the same eigenvalues.