Stability of $\dot{x}=x^3-2xy,\dot{y}=x^2-y$ I was given the following exercise.

Consider the nonlinear dynamical system 
$\dot{x}=x^3-2xy, \dot{y} = x^2-y$
(a) Find all fixed points.
(b) Determine whether each fixed point is linearly stable or not and draw its local phase portrait.

I am not sure about (b).
My attempt:
(a)
To find the fixed points, we need to solve $x^3-2xy=0,x^2-y=0$. The first equation is equivalent to $x(x^2-2y)=0$, so we have $x=0$ or $x^2-2y=0$. Substituting $x=0$ into $x^2-y=0$ yields $y=0$, hence we have a fixed point $(0,0)$. Now, substituting $x^2-2y=0$ into $x^2-y=0$ yields $y=2y$, so $y=0$. Substitute back and we get $x=0$. Again we get a fixed point $(0,0)$. Therefore the fixed point is $(0,0)$.
(b)
Let $f=x^3-2xy, g=x^2-y$. Then we have $f_x=3x^2-2y, f_y=-2x, g_x=2x, g_y=-1$. Therefore the Jacobian at $(0,0)$ is $\begin{pmatrix} 0 & 0 \\ 0 & -1 \end{pmatrix}$. We can easily see that the eigenvalues of this matrix is $0$ and $-1$, and the corresponding eigenvectors are $(1,0), (0,1)$, respectively.
Now here's the question. In the lecture, nothing was talked about fixed points with zero eigenvalues. So I am not really sure how to proceed.
By naively applying the linearization, we have a linear approximation of the form $Av_1e^{0t}+Bv_2e^{-1t}=Av_1+Bv_2e^{-t}$ where $A, B$ are some constant and $v_1=(1,0)^t$ and $v_2=(0,1)^t$ are eigenvectors. Since $t\to \infty$ does not yield the origin, I felt like this is not stable(because a stable point is like a black hole, right?), but I am not really sure.
Also, I am not sure how to draw a phase plane.
 A: Let's proceed step by step. There are some issues with your solution
Finding the fixed points:
From $\dot{x}=0$ we have that either $x=0$ or $y=\frac{1}{2}x^{2}$.
From $\dot{y}=0$ we have that $y=x^{2}$.
So $(0,0)$ is the only fixed point.
Stability
The Jacobian at (0,0) is
$$
J=\begin{bmatrix}
0&0\\0&-1\end{bmatrix}
$$
Which has eigenvalues $\lambda_{1}=0$ and $\lambda_{2}=-1$. Recall that when solving a system of linear (in this case linearized) ODEs you are basically expressing the variables as the sum of two (orthogonal) components associated with those eigenvalues. The interpretation of the 'second component' is easy: it decays as $t\to\infty$. Now the first one is trickier, but notice that since $e^{\lambda_{1}t}=1$ it means that it will stay where it started and don't 'move'.
Now the Jacobian is not very useful to study the dynamics near the fixed point in this case because one of the eigenvalues is zero. Thus, we cannot apply the Hartman-Grobman theorem and the linearized will not be a good approximation in a neighbourhood.
However, we can try to interpret the findings by assuming our system was linear. If the true system was linear then $\lambda_{1}=0$ means that the system is not stable: the 'first component' will remain in the initial condition, and by construction will not converge unless it starts at the fixed point. This means that if we want our system to converge to the steady state we need to 'kill' the first component in our system of ODEs, so the system is saddle-path stable. We will find that this is indeed true for the nonlinear system as well but this will require some more work.
Phase Diagram
We can study the dynamics as follows.
Consider $\dot{x}=x\left(x^{2}-2y\right)$. We know that if $x=0$ of $y=\frac{1}{2}x^{2}$ then $\dot{x}=0$. In addition, you can study what happens 'off path'.

*

*Consider the area above the parabola, that is $y>\frac{1}{2}x^{2}$. If $x>0$ then $\dot{x}<0$ and if $x<0$ then $\dot{x}>0$.

*Consider the area below the parabola, that is $y<\frac{1}{2}x^{2}$. If $x>0$ then $\dot{x}>0$ and if $x<0$ then $\dot{x}<0$.

Now consider $\dot{y}=x^2-y$. We know that if $y=x^{2}$ then $\dot{y}=0$. Now if $y>x^{2}$ then $\dot{y}<0$ and if $y<x^{2}$ then $\dot{y}>0$. That is, $y$ grows outside the parabola and decreases inside the parabola.
The phase diagram basically consists in graphing everything and draw some arrows to indicate how $x$ and $y$ evolve off $\dot{x}=0$ and $\dot{y}=0$ (I'm sorry for my bad drawing skills)

As you can notice graphically, you can converge to (0,0) from above and from some trajectories (I don't think all of them). I took the liberty of plotting the phase plane together with some trajectories to verify this

A: From the wikipedia article:

If all the eigenvalues have negative real part, then the solution is
called linearly stable.

Thus, if there is a zero eigenvalue, then the system is not linearly stable.
The phase portrait is shown in the picture:

As we can see, the equilibrium point is asymptotically stable.
