Show that $(x_1,x_2)=(0,0)$ is an unstable fixed point for the system $$\dot{x_1} = -x_1 + x_2^6\qquad\dot{x_2} =x_2^3 + x_1^6$$ Hint: Consider the Lyapunov function $V(x_1,x_2) = ax_1^i + bx_2^j$.
What I have done: Let $$f(x):={-x_1 + x_2^6 \choose x_2^3 + x_1^6}$$ then $$\dot V = \nabla V\cdot f(x) = (aix_1^{i-1},bjx_2^{j-1})\cdot{-x_1 + x_2^6 \choose x_2^3 + x_1^6} = aix_1^{i-1}(x_2^6 - x_1) + bjx_2^{j-1}(x_1^6+x_2^3).$$
I have a theorem that basically says that if $\dot V(x) > 0$ for all $x\in E \setminus \{(0,0)\}$, where $E$ is an open subset of $\mathbb R^2$ containing $(0,0)$, then $(0,0)$ is unstable.
Here's my main question: We were given the hint to consider the given Lyapunov function, but I'm having some difficulty understanding how to use this Lyapunov function to show that basically any nonzero $x\in\mathbb R^2$ gives us that $\dot V(x) > 0$. Must I remain general with the $i$ and $j$, or should I arbitrarily pick $i$ and $j$ in this case? If we don't, then we'd have to break this up into cases, but from what I understand, we just need to find a function that satisfies the criteria for the Lyapunov function, not show it works for every possibility.