This is essentially trying to prove Liouville's theorem holding for time-dependent systems. Where S is the action coming from the Hamiltonian system.
$\det\frac{dS^t}{dx} = 1+ (\nabla f)*t + 0(t^2)$ where t is evaluated at zero for $\dot{x}=f(x,t)$
I have proven $1+\operatorname{tr}(A)*t+0(t^2)$ which I'm assuming is the key part to this proof.
Where I was heading with this is after I have proven $1+\operatorname{tr}(A)*t+0(t^2)$ is, I could say the time-dependent system is being evaluated at $t=0$ so then the $\operatorname{tr}(A)$ is just that $(\nabla f)$ part.