Sorry to be sloppy the first time. Now I have corrected something according to the comments.
Let's consider this system of $n$ linear second order ODEs: $$\ddot {\mathbf{x}}=f(\mathbf{x}),$$ Where $f:\mathbb R^n\to \mathbb R^n$, and $f(\mathbf 0)=\mathbf 0$. This is the equation of motion of an oscillator with stable equilibrium $\mathbf 0$ (not necessarily simple harmonic).
Improtant: $\mathbf 0$ is a stable equilibrium.
We can approximate this by $\ddot {\mathbf{x}}=A \mathbf{x}$, where $A=Df_{\mathbf{x}=\mathbf 0}$. Intuitively this shoud lead to sinusoidal waveform. However, I struggle to prove this mathematically. To prove that this leads to a sinusoidal waveform, I just need to prove that all eigenvalues of $A$ are negative OR zero. Now I have realised that the converse is not true, so just focus on the forward direction.
My attempt: Informally, if the equilibrium is stable, forces should be roughly "directed towards it". $f(\mathbf x)=A\mathbf x$ for small $\mathbf x$, so this translates to, $$ \ddot{\mathbf x}\approx -k\mathbf x,k\leq0\\ A\mathbf x\approx -k\mathbf x\\ \mathbf x^TA\mathbf x\approx -k|\mathbf x|^2. $$ So I guess $A$ should be negative definite...or something like that.
How can I prove that the eignevalues are all negative OR zero?
Here is a possibility... Assume that $A$ has a positive eigenvalue $\lambda > 0$, so $Av = \lambda v$ for some $v \in \mathbb{R}^n$. If ${\bf 0}$ is a stable equilibrium point, then there exists an $\epsilon > 0$ such that if the initial condition starts within the closed $\epsilon$-ball about ${\bf 0}$, it will remain in that $\epsilon$-ball. Suppose that $x(0) = {\bf 0} + \epsilon v$ and $x'(0) = 0$. Then it suffices to show that $||x(t) - {\bf 0}|| > \epsilon$ at some time $t > 0$.
Looking at the linear approximation of $f$ about the origin, the approximate differential equation is $x''(t) = Ax(t)$, and so initially we have $x''(0) = \epsilon Av = \epsilon \lambda v$, meaning that it will have positive acceleration in the $v$ direction.
Sure, there are minor details omitted, but hopefully you understand the argument as to why the derivative matrix can not have a positive eigenvalue?