Show that the following is asymptotically stable - Liapunov function I am trying to figure out how to show that the zero solution to the following is asymptotically stable.
$$x'=xy-x^3$$
$$y'=-y+x^3y$$
I was trying to use a liapunov function: 
$$V(x,y)=\frac{1}{2}[x^2+y^2]$$
Which is always positive. This gives us that 
$$V'(x,y)=xx'+yy'$$
$$=x(xy-x^3)+y(-y+x^3y)$$
$$=x^2y-x^4-y^2+x^3y^2$$
I know that I need this to be strictly negative for some domain surrounding and including the origin. I am having a hard time showing this. I have tried a couple other liapunov functions as well, but none have been successful either.
Thanks for any help! I am also open to other methods if there is something easier. When I linearized the system though, it seemed to be inconclusive.
I found the linearization to be 
$$
\begin{bmatrix}
    x'\\
    y'\\
\end{bmatrix}=
\begin{bmatrix}
    y && -\frac{x^3}{y}\\
-\frac{y}{x} && x^3 \\
\end{bmatrix}
\begin{bmatrix}
    x\\
    y\\
\end{bmatrix}
$$
So I can't plug in $x=y=0$ to find the eigenvalues. Unless I am not doing that right...
 A: This is an example of the interesting case of a null eigenvalue, aka a non hyperbolic critical point :
The Jacobian (aka the Linearisation matrix) of the given system is  :
$$J(x,y) = \begin{bmatrix} y-3x^2 & x \\ 3x^2y & -1+x^3\end{bmatrix}$$
The linearisation matrix for the point $(x,y) = (0,0)$ is :
$$J(0,0) = \begin{bmatrix} 0 & 0 \\ 0 & -1 \end{bmatrix}$$
We yield the eigenvalues :
$$\det(J(0,0) - \lambda I) = \begin{vmatrix} -\lambda & 0 \\ 0 & -1-\lambda \end{vmatrix}=0\Rightarrow\lambda(1+\lambda)=0 \Rightarrow \lambda =\begin{cases} 0 \\ -1\end{cases}$$
We notice that we have a null eigenvalue and also $\det(J(0,0)) = 0$, thus the origin $(x,y) = (0,0)$ is a non hyperbolic critical point.
We transform our system on the form :
$$\begin{bmatrix} x \\ y \end{bmatrix}' = \begin{bmatrix} 0 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} + f(x,y)$$
where $f(x,y)$ is a function, such $f(x,y) = (xy-x^3,x^3y) : \mathbb R^2 \to \mathbb R^2$.
Solving the equation :
$$-y(x) + f_2(x,y(x)) = 0 \Rightarrow -y(x) + x^3y(x) = 0 \Rightarrow y(x)(x^3-1)=0 \implies y(x) = 0$$
Finally : 
$$f_1(x,y(x)) = f_1(x,0) = -x^3$$
The function $f_1(x,y(x))$ that we yielded is of the form $f_1(x,y(x)) = ax^k + \mathcal{O}(|x|^{k+1})$, since there is a continuous function $g(x)$ such that $g(x) = \mathcal{O}(|x|^{k+1})$ with $k=3$ and $g(x) = 0 \cdot |x|^4$. 
Since $f_1(x,y(x)) = -x^3$, the coefficient of the $x$ order is a negative number and the power $k=3$ is an odd number. Thus, the origin $(x,y) = (0,0)$ is an asymptotically stable critical point for the given system.
Drawing a stream plot (phase portrait) for the given dynamical system, we also observe the asymptotic stability of the origin $(x,y) = (0,0)$ :
$\qquad \qquad \qquad \qquad \qquad \quad$  
