Using a Lyapunov function to classify stability and sketching a phase portrait 
Consider the system
  $$x' = -x^3-xy^{2k}$$
$$y' = -y^3-x^{2k}y$$
  Where $k$ is a given positive integer.
a.) Find and classify according to stability the equilibrium solutions.
$\it{Hint:}$ Let $V(x,y) = x^2 + y^2$
b.) Sketch a phase portrait when $k = 1$
$\it{Hint:}$ What are $x'$ and $y'$ when $y=ax$ for some real number $a$?

a.)
Using $V$, we get $\frac{d}{dt}V=2xx'+2yy'$
Plugging in our system , we get:
$$\frac{d}{dt}V=2x(-x^3-xy^{2k})+2y(-y^3-x^{2k}y)$$
$$=-(x^4+y^4)-x^2y^{2k}-x^{2k}y^2<0$$
I dropped the $2$ since it doesn't matter to determine stability. We see that our own equilibrium is $(0,0)$ since setting $x'=0$ we get
$$y^{2k}=-x^2$$
Which only works for $x=y=0$
Therefore our system is asymptotically stable at the origin.
I am having trouble with b.), mostly because the hint is confusing me.
Let $y=ax$, then our system becomes
$$x'=-x^3-a^2x^3=-x^3(1+a^2)$$
$$y'=-a^3x^3-ax^3=-ax^3(1+a^2)$$
I am not sure what to do with this. Using linearization doesn't work since the Jacobian will be the zero vector at the point of interest. I have never had a problem that asks to draw a phase portrait when linearization doesn't work, so I am hoping someone more clever than me can offer some advice.
 A: Although there are many ways to do this, I suspect what the problem is guiding you towards doing is to obtain the flow directly by evaluating over every line that intersects the origin in phase space.
So for a sketch, you would draw the line $y = 0.1 x$, and use the expression you found above for $a = 0.1$ to determine the magnitude and direction of the flow on that line. Then try it for a couple of other different lines, and use common sense to fill in the rest.
A: Phase portraits - a partial offering
Below are phase portraits for $k=1,2,5$. The red lines indicate the null clines where $\dot{y}=0$ and $\dot{y}=0$.
$k = 1$
The linear system is
$$\begin{align} 
\begin{split} 
\dot{x} &= -x^{3} - xy^{2} = -x \left( x^{2} + y^{2} \right) \\ 
\dot{y} &= -y^{3} - x^{2}y = -y \left( x^{2} + y^{2} \right) 
\end{split} 
\end{align}$$
$$ \dot{r} = \frac{x \dot{x} + y \dot{y}}{r} = -r^{3} $$
The lone critical point is the origin.
When $y = a x$, $a\in\mathbb{R}$, we have
$$\begin{align} 
\begin{split} 
\dot{x} &= -x^{3}\left( 1 + a^{2} \right)  \\ 
\dot{y} &= -a y^{3}\left( 1 + a^{2} \right)
\end{split} 
\end{align}$$

$k = 2$
$$\begin{align} 
\begin{split} 
\dot{x} &= -x^{3} - xy^{4} = -x \left( x^{2} + y^{4} \right) \\ 
\dot{y} &= -y^{3} - x^{4}y = -y \left( x^{2} + y^{2} \right) 
\end{split} 
\end{align}$$
$$ \dot{r} = \tfrac{1}{8} r^3 \left(\left(r^2-2\right) \cos (4 \theta )-r^2-6\right) $$
The bounding curves for $\dot{r}$ are when $\cos 4\theta = 1$
$$\dot{r} = -r^{3}$$
and when $\cos 4\theta = -1$
$$\dot{r} = -\tfrac{1}{4} r^3 \left(r^2+2\right)$$
The bounding curves cross at $r=\sqrt{2}$. At no point is $\dot{r}$ ever positive.


