Suppose we have the differential system of the form,
$\dot{x}_1 = -x_1 x_2$
$\dot{x}_2 = \omega(1-a x_2)$, where $\omega > 0$ and $a > 0$ are positive constants.
How can I prove that the system is globally asymptotically stable in $x_1^* = 0$ and $x_2^* = a^{-1}$ via the Lyapunov method?
I have tried with the Lyapunov candidate function $V = 0.5 x_1^2 + 0.5 (x_2 - a^{-1})^2$, but I didn't get far.
Let's see if this logical argument can be applied to your case. Your original Lyapunov function candidate can be applied, with a little modification.
To apply Lyapunov's theorem, the coordinate can be changed to
$$ \epsilon = y - a^{-1} $$
and the system becomes
$$ \dot{\epsilon} = - \omega a \epsilon $$
$$ \dot{x} = - x (\epsilon + a^{-1}) $$
Since $\dot{\epsilon}$ doesn't depend on $x$, then the solution for the linear time-invariant ODE $\dot{\epsilon} = - \omega a \epsilon$ is
$$\epsilon(t) = \epsilon_{0} e^{-\omega a t} $$
and it is clear that $\epsilon(t)$ exponentially converges to $0$. As a result, we can rewrite $\dot{x}$ as a time-variant ODE
$$ \dot{x} = - (\epsilon_{0} e^{-\omega a t} + a^{-1}) x .$$
Next, we have the Lyapunov function
$$ V = \frac{1}{2} x^{2} $$
and the time derivative of $V$ is
$$ \dot{V} = - (\epsilon_{0} e^{-\omega a t} + a^{-1}) x^{2} .$$
Since $\lim_{t \rightarrow \infty} \epsilon_{0} e^{-\omega a t} = 0$, then $$ \lim_{t \rightarrow \infty} \dot{V}(t) = - (a^{-1}) x^{2} .$$
Thus, after some time $t$,
$$ \dot{V} \leq - (a^{-1}) x^{2} < 0 \; \text{for } x \neq 0.$$