Lyapunov Functions and Basins of Attraction \begin{align}
    x' &= -x^3 + x^5 + (x^4)(y^5)\\[.7em]
    y' &= -8y^3 + y^5 - 10(y^4)(x^5)
\end{align}
$(0,0)$ is obviously a critical point of the system, and we are given that it is asymptotically stable, but have to show it. 
I have tried to make a Lyapunov function $V(x,y) = ax^2 + cy^2$, with a,c > 0 but I am having trouble to prove that $\frac{d}{dt} V(x,y)$ is negative definite. I get some complicated polynomial I can't use logic to finalize. How can I change the Lyapunov to come up with a meaningful conclusion?
\begin{align}
    \frac{d}{dt}V(x,y) = 2ax(-3x^2 + 5x^4 + 4x^3y^5) + 2cy(-24y^2+5y^4-40y^3x^5)\\[.7em]
\end{align}

Hint: choose $a$ and $c$ so the terms in $x^5 y^5$ cancel.  Note that for $x$ near $0$, lower powers of $x$ dominate higher powers, and similarly for $y$.
