Question about equilibrium points of non-linear system of ODEs. So I have the system
\begin{align}
x'&=y\\
y'&=-x-y\ln(x^2+4y^2)
\end{align}
To find the equilibrium points I need $x'=0$ and $y'=0$, thus I obtain
\begin{align}
y&=0\\
-x-y\ln(x^2+4y^2)&=0
\end{align}
I don't see how to proceed here. If $y=0$ in the second equation we get $-x=0\Leftrightarrow x=0$ but if $x=0$ and $y=0$ the logarithm is undefined. So $(0,0)$ can't be an equilibrium point.

Let $V((x,y)) = {1 \over 2} (x^2+y^2)$, and $\phi(t) = V((x(t),y(t)))$. Note that
$\phi'(t) = -y^2 \ln(x^2+4y^2)$.
Let $A= \{ (x,y) | {1 \over 4} \le x^2+y^2 \le 4 \}$.
Note that $\phi'(t) \ge 0$ if $x(t)^2+y(t)^2= {1 \over 4}$.
Note that $\phi'(t) \le 0$ if $x(t)^2+y(t)^2= 4$.
$A$ contains no equilibrium points.
