Prove that $(0,0)$ is asymptotically stable and that its basin of attraction is $ \mathbb{R}^2 $. Consider a differential equation $ (x',y')=F(x,y) $, where $ F:\mathbb{R}^2\to \mathbb{R}^2 $ is $ C^1 $ and $ F(0,0)=(0,0) $. Let $ V:\mathbb{R}^2\to \mathbb{R} $ be a strict Lyapunov function associated to $ (0,0) $. I want to prove that if $$ \lim\limits_{\|(x,y)\|\to +\infty}V(x,y)=+\infty $$ then $ (0,0) $ is asymptotically stable and its basin of attraction is $ \mathbb{R}^2 $. I believe I have to prove that $V'<0$ for all $(x,y)\neq (0,0)$, but how can I prove that when I don't know $F(x,y)$? Furthermore, what does it mean that its basin of attraction is $\mathbb{R}^2$? That $(0,0)$ is a point of attraction for all of $\mathbb{R}^2$ so that every flow curve will terminate at $(0,0)$? Finally, what does the word "strict" indicate?

The definition of a strict Lyapunov function and the chain rule tell you that for any solution $(x,y) = P(t)$, $ \dfrac{d}{dt} V(P(t)) \le 0$, with equality only when $P(t) = (0,0)$.

If you had a solution $P(t)$ that was not attracted to $(0,0)$, it would have to stay in some annulus $\\{(x,y): r^2 \le x^2 + y^2 \le R^2\\}$ for all $t \ge 0$. Now use the fact that a continuous function on a compact set attains a minimum.