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.
I shall provide an intuitive method to sketch the phase portraits which I have learnt back at university years ago. In simpler terms, it shows the trajectory of your gradient field. It is relatively straightforward given a first-order $2 \times 2$ matrix linear ODE system. We shall leverage on the info you have provided above and some Linear Algebra to analyze the dynamics of the system.
We can rewrite our system of linear ODE in matrix form.
$$
\begin{bmatrix}
\dot x \\
\dot y \\
\end{bmatrix} =
\begin{bmatrix}
0 & -1 \\
1 & a \\
\end{bmatrix}
\begin{bmatrix}
x \\
y \\
\end{bmatrix}
$$
$$
\implies A =
\begin{bmatrix}
a_{11} & a_{12} \\
a_{21} & a_{22} \\
\end{bmatrix} =
\begin{bmatrix}
0 & -1 \\
1 & a \\
\end{bmatrix}
$$
For a $2 \times 2$ system, the characteristic polynomial can be given by:
$$ p(\lambda) = \lambda^2-\text{tr}(A)\lambda + \text{det}(A) $$
The roots or the eigenvalues to the characteristic polynomial is given by setting $p(\lambda)=0$. Then:
$$ \lambda_{1,2}=\frac{\text{tr}(A)}{2} \pm \sqrt{\Big(\frac{\text{tr}(A)}{2}\Big)^2-\text{det}(A)} $$
If $A$ is diagonalized, ie. $A=Q\Lambda Q^{-1}$, where $Q$ consists of eigenvectors $q_i$ and eigenvalues in $\Lambda$, we can find our eigenvectors by computing:
$$q_i =
\begin{bmatrix}
-a_{12} \\
a_{11} - \lambda_i \\
\end{bmatrix}
$$
The eigenvectors will show the trajectory of the phase portrait.
From the discriminant $\Delta$ and $a > 0$, we will have to analyze the system for $a \in (0,2),$ $a \in (2,\infty),$ and $a =2$. I will skip the computations and present the concept directly. I leave it to you to DIY.
Case $1$: $a \in (0,2)$. This implies $\Delta<0$. The eigenvalues are complex. For simplicity, let us assume $a=1$.
$$\lambda_{1,2} = \frac{1}{2} \pm i \frac{\sqrt3}{2} $$
For this case, we shall just consider one of the eigenvectors, for the corresponding eigenvalue has a positive square-root. Its real and imaginary parts of the (complex) eigenvector will determine the trajectory of the field.
$$q_1 =
\begin{bmatrix}
-(-1) \\
0 - \big( \frac{1}{2} + i\frac{\sqrt3}{2} \big) \\
\end{bmatrix} =
\begin{bmatrix}
1 \\
- \frac{1}{2} - i\frac{\sqrt3}{2} \\
\end{bmatrix}
$$
$$\implies q_{\text{re}} =
\begin{bmatrix}
1 \\
- \frac{1}{2}\\
\end{bmatrix}, -q_{\text{im}}
\begin{bmatrix}
0 \\
\frac{\sqrt 3}{2}\\
\end{bmatrix}
$$
As the eigenvalue is complex with $\text{Re}\{\lambda_{1,2}\} > 0$, it is an unstable focus. We shall sketch $q_{\text{re}}$ and $-q_{\text{im}}$ in the phase plane as follows.
Case $2$: $a \in (2,\infty)$. This implies $\Delta > 0$. The eigenvalues are real and not the same. For simplicity, let us assume $a=3$.
$$ \lambda_1 = \frac{3}{2} + \frac{\sqrt{5}}{2} = \frac{1}{2}(3 + \sqrt5) \implies
q_1 =
\begin{bmatrix}
-(-1) \\
0 - \frac{1}{2}(3 + \sqrt5)\\
\end{bmatrix} =
\begin{bmatrix}
1 \\
- \frac{1}{2}(3 + \sqrt5)\\
\end{bmatrix}
$$
$$ \lambda_2 = \frac{3}{2} - \frac{\sqrt{5}}{2} = \frac{1}{2}(3 - \sqrt5) \implies
q_2 =
\begin{bmatrix}
-(-1) \\
0 - \frac{1}{2}(3 - \sqrt5)\\
\end{bmatrix} =
\begin{bmatrix}
1 \\
- \frac{1}{2}(3 - \sqrt5)\\
\end{bmatrix}
$$
$$\lambda_i: \lambda_1 > \lambda_2 > 0 $$
The eigenvalue with the highest magnitude, i.e. $|\lambda_i|$ is known as the 'fast' eigenvalue. The trajectory will take the direction of the eigenvector corresponding to the 'fast' eigenvalue as it converges to (or diverges from) the origin. Since $\lambda_1 > \lambda_2 > 0$, it is an unstable node.
Case $3$: $a = 2$. This implies $\Delta = 0$. The eigenvalues are real and the same (repeated). This shows a degenerate system. This also means $A$ is not diagonalizable. The trajectory take straight lines through the origin. We can deduce the trajectory by multiplying $A$ with the standard basis vectors as follows:
$$
\lambda_1 = \lambda_2 = \frac{2}{2} = 1
$$
$$
q_1 =
\begin{bmatrix}
0 & -1 \\
1 & 2 \\
\end{bmatrix}
\begin{bmatrix}
1 \\
0 \\
\end{bmatrix} =
\begin{bmatrix}
0 \\
1 \\
\end{bmatrix}
$$
$$
q_2 =
\begin{bmatrix}
0 & -1 \\
1 & 2 \\
\end{bmatrix}
\begin{bmatrix}
0 \\
1 \\
\end{bmatrix} =
\begin{bmatrix}
-1 \\
2 \\
\end{bmatrix}
$$
Since $\lambda_1 = \lambda_2 > 0$, it shows an unstable degenerate node.
Alternatively, you can check out this video on how to sketch phase portraits: https://www.youtube.com/watch?v=dpbRUQ-5YWc
Hope it helps. Cheers!
Best Answer
No, it is not a gradient field. In particular, it is not the gradient of the energy $E$. On the contrary: the energy $E$ is conserved. Thus the trajectories are the curves of constant $E$.