Let $x'=f(x), x'=g(x)$ be two ODE, with $f(x),g(x):\mathbb R^2\rightarrow\mathbb R^2$, such that $\langle f(x),g(x)\rangle =0$ for all $x\in\mathbb R^2$. If $f$ has a periodic orbit then $g$ has equilibrium point (i.e., exists $x_0$ such that $g(x_0)=0$).
So far I just got that if $x(t)$ is a periodic orbit of $f$, with period $t_0$, then $x(t)=x(t+t_0)\Rightarrow x'(t)=x'(t+t_0)\Rightarrow f(x(t))=f(x(t+t_0))$, so $\langle f(x(t),g(x(t))\rangle = \langle f(x(t+t_0),g(x) \rangle=\langle f(x(t+t_0),g(x(t+t_0))\rangle$, then $g(x)=kg(x(t+t_0))$, for one $k\in\mathbb R$.
I can't see how to go on, is this the right path? Or there is a better way to do?
Thanks in advance.
You have to know index theory for ODE in a plane. Suppose that $x'=f(x(t))$ has a periodic orbit $x(t)$ and for every $t$, $g(x(t))\neq 0$, since $\langle f(x(t)),g(x(t))\rangle=0$, you deduce that for every $t$, $g(x(t))$ points outward of the curve $x(t)$ or for every $t$, $g(x(t))$ points inward of the curve $x(t)$. You can then apply index theory (see section 6 p.8 of the reference) to deduce that $g$ has a zero in the region bounded by $x(t)$.
https://www.math.psu.edu/bressan/PSPDF/M417-review3.pdf