Lyapunov stability of $\begin{cases}x' = e^{ay} - e^x \\ y' = a x^2 + (a-a^2) x + ay^2 e^{-y}\end{cases}$

Consider the system:

$$\begin{cases}x' = e^{ay} - e^x \\ y' = a x^2 + (a-a^2) x + ay^2 e^{-y}\end{cases}$$

Using Lyapunov first method we have, for $$a \in ]-\infty,1[ \cup ]0,1[$$ $$p = (0,0)$$ is unestable and for $$a \in ]1,+\infty[$$ is asymptotically stable.

For $$a = 1$$ I need to use Chetaev theorem to show that it is unestable and without indication I have to determine the stability when $$a = 0$$.

$$a = 1$$

In this case I looked for a Lyapunov function with separated variables but I get:

$$\dot V(x,y) = V_1'(x) e^y - V_1'(x) e^x + x^2 V_2'(y) + y^2 e^{-y}V_2'(y)$$

I don't see how to choose $$V_1,V_2$$ to make $$\dot V(x,y) > 0$$ in a neighbourhood of $$p = (0,0)$$

$$a = 0$$

This time we have the system $$\begin{cases}x' = 1 - e^x \\ y' = 0\end{cases}$$ and one solution is $$(0,C)$$ with $$C \in \mathbb{R}$$, so $$p = (0,0)$$ is not asymptotically stable. I tried to use Lyapunov's second theorem to have $$\dot V = 0$$ and show stability but it didn't work again.

The Chetaev function for $$a=1$$ is $$V(x,y)=y$$, since for any $$(x,y):\; y>0$$ we have $$\dot V= x^2+y^2e^{-y}>0$$.
For the case $$a=0$$, consider the Lyapunov function $$V(x,y)=e^x-x-1+y^2$$: $$\dot V= e^x\dot x-\dot x= -(e^x-1)^2\le 0.$$