A dynamical system \begin{equation} \dot x = f(x,y) \quad\quad \dot y = g(x,y) \end{equation} has a conserved quantity $E$ if \begin{equation} \frac{dE}{dt} = \frac{\partial E}{\partial x}\dot x + \frac{\partial E}{\partial y}\dot y = 0 \end{equation} I'm trying to understand the conditions on $f$ and $g$ for such an $E$ to exist.
My understanding is that this depends on the equation \begin{equation} \frac{dy}{dx} = \frac{g(x,y)}{f(x,y)} \end{equation} being exact, i.e. \begin{equation} \frac{\partial g}{\partial y} = -\frac{\partial f}{\partial x} \end{equation} in which case the equation can be integrated to $E$. Thus, for instance, for the simple harmonic oscillator \begin{equation} \dot x = y \quad\quad \dot y = -\omega^2x \end{equation} we have \begin{equation} \frac{\partial g}{\partial y} = 0 = -\frac{\partial f}{\partial x} \end{equation} So a conserved quantity exists, obtained from \begin{equation} y\frac{dy}{dx} + \omega^2x = 0 \end{equation} which integrates to the usual \begin{equation} E = \frac12\omega^2x^2+\frac12y^2 \end{equation} However, this doesn't work for a different example (from https://services.math.duke.edu/~jtwong/math356-2019/lectures/Dynamics4_cons_lyapunov.pdf): \begin{equation} \dot x = x(1-y) \quad\quad \dot y = ay(x-1) \end{equation} That is, \begin{equation} x(1-y)\frac{dy}{dx} + ay(1-x) = 0 \end{equation} is not exact. However, we can multiply through by $1/xy$ to get \begin{equation} \frac{1-y}y\frac{dy}{dx} + a\frac{1-x}x = 0 \end{equation} which is exact.
This leads me to think that the condition on $f$ and $g$ for a conserved quantity to exist is that there exists some function $a(x,y)$ such that \begin{equation} a(x,y)f(x,y)\frac{dy}{dx} - a(x,y)g(x,y) = 0 \end{equation} is exact. When will this not be the case?
Moreover, in the case of a 2nd order ODE \begin{equation} \dot x = y \quad\quad \dot y = g(x,y) \end{equation} how does this proposition impose $g(x,y)=g(x)$, which I believe it should?
(Sorry for the length.)