The full problem:
Determine the qualitative behavior near the non-hyperbolic critical point at the origin for the system with $\dot{x} = xy$ and $\dot{y} = -y-x^2$. Sketch the phase portrait.
I have what I think is a correct answer, I just wanted to check the correctness of my statement. I also have a phase portrait, but I double checked it's correctness in some software.
My answer:
First off, here it is easy to tell that the only equilibrium point we have here is indeed the origin. Next, let's discuss the behavior of this solution point.
$$Df(x,y) = \left( \begin{matrix}y & x\\ -2x & -1\end{matrix}\right)$$ $$Df(0,0) = \left( \begin{matrix}0 & 0\\ 0 & -1\end{matrix}\right)$$
We get then that our eigenvalues are $\lambda_1 = 0$ and $\lambda_2 = -1$. This shows us that the point is indeed stable, and will behave similarly to a stable node. In other words, any solution will sink towards the origin, however, when approaching from $y > 0$, the trajectory will move around the origin because of the first zero eigenvalue.
Your answer is incorrect.
The fact that the eigenvalues are $0$ and $-1$ does not tell you whether the critical point is stable or unstable: it could be either. You need more information.
For example, $\dot{x} = x^2,\; \dot{y} = -y$ also has eigenvalues $0$ and $-1$ but has an unstable critical point.