The points $(\epsilon,0)$ are themselves equilibria, so clearly $(0,0)$ is not asymptotically stable.
If you want to investigate the local stability of an equilibrium point of a nonlinear system it is usually easier to linearize. If all eigenvalues of linearized system have a negative real part then that equilibrium point is (asymptotically and exponentially) stable and if any of the eigenvalues has a positive real part it is unstable. Only in the edge case, when it is not unstable but there are eigenvalues with zero real part, would it be required to consider more of the nonlinear dynamics in order to identify whether or not the equilibrium point is locally stable.
It can be noted that if you are able to show that the system is stable near the equilibrium point $x_{eq}$ by using the linearization
$$
\dot{e} \approx A\,e
$$
with $e = x - x_{eq}$ and $\|e\|$ close to zero, you can also always find a Lyapunov function by solving the (continuous) Lyapunov equation
$$
A^\top P + P\,A = -Q,
$$
for any given positive definite $Q=Q^\top$ the solution for $P$ is also positive definite. Namely, the associated Lyapunov function is given by $V(e) = e^\top P\,e$, for which it holds that $\dot{V}(e) \approx -e^\top Q\,e$ when $\|e\|$ is close to zero. A key thing to note here is that in general it is not true that $\dot{V}(e) \nleq 0\ \forall\, e$. Namely, if it would hold for all $e$ then you would have shown global instead of local stability. One final note is that the set of all $e$ such that $\dot{V}(e) < 0$ is a lower bound of the basin of attraction of that equilibrium point.
Best Answer
The origin is unstable which can be shown with the Chetaev instability theorem.
Let $ q = [x, y]^T $ and $ V(q) = xy $. Trivially, $ V(0) = 0 $ and $ V(q) > 0 $ in the first quadrant. Since $ \dot{V}(q) = x^4 + y^4 > 0 $ for all $ x,y \in \mathbb{R} \setminus \{0\} $, the origin $ q = 0 $ is unstable from Chetaev instability theorem. Let me state the theorem (see Theorem 4.3 in Hassan K. Khalil, Nonlinear Systems, 3rd Edition)
Theorem: Let $ q = 0 $ be an equilibrium point for $ \dot{q} = f(q) $. Let $ V: D \rightarrow \mathbb{R} $ be a continuously differentiable function such that $ V(0) = 0 $ and $ V(q_0) > 0 $ for some $ q_0 $ with arbitrarily small $ \Vert q_0 \Vert $. Define a set $ U $ as $ U = \{ q \in B_r ~|~V(x) > 0 \} $ and suppose that $ \dot{V}(q) > 0 $ in $ U $. Then, $ q = 0 $ is unstable.