How are the points in an elliptic curve over the finite field calculated? For the elliptic curve with equation $y^2 = x^3 + x + 0 \pmod{13}$ in the finite field, how are all the points in this curve calculated. See the image below for an example created via https://graui.de/code/elliptic2/.

The above images shows 20 points. How are the coordinates of these points calculated?
Given
$$
y^2 = x^3 + x \pmod{13}\quad (1)
$$
note that in addition to the point at infinity, the other points are the solutions $(x,y) \in \mathbb{F}_{13}\times  \mathbb{F}_{13}$ of (1). With algebraic construction we add the point of infinity ($\{\mathcal O\}$) to the set of the solutions $$E := \{ (x, y) \in (\mathbb{F}_{13})^2 \mid y^2 = x^3 + x \} \cup \{\mathcal O\}.$$
Since you need to solve (1) for $y^2$ this implies that whenever $(x,y)$ is a solution so is $(x,-y).$ For a small field like this list the squares
$$\begin{array}{c|ccccccc}\hline
y &amp; 0 &amp; \pm 1 &amp; \pm 2 &amp; \pm 3 &amp; \pm 4 &amp; \pm 5
&amp;\pm 6\\ \hline
y^2 &amp; 0 &amp; 1 &amp; 4 &amp; 9 &amp; 3 &amp; 12 &amp; 10\\ \hline
\end{array}$$
and note that everything must be reduced mod 13. So $-4$ is actually $9.$
Now, whenever $x^3+x$ is equal to $y^2$ we pick up $(x,y),(x,-y)$ as points on the curve. So now prepare the table
$$\begin{array}{c|ccccccccccccccc}\hline
x &amp; 0 &amp; 1 &amp; 2 &amp; 3 &amp; 4 &amp; 5 &amp; 6 &amp;7 &amp;8 &amp;9 &amp;10 &amp;11 &amp;12\\ \hline
x^3 &amp; 0 &amp; 1 &amp; 8 &amp; 1 &amp; 12 &amp; 8&amp; 8&amp; 5&amp; 5&amp; 1&amp; 12&amp; 5&amp; 
12\\ \hline
x^3+x &amp; 0^* &amp; 2 &amp; 10^* &amp; 4^* &amp; 3^* &amp; 0^* &amp; 1^* &amp; 12 &amp; 0^*&amp; 10^*&amp;  9&amp; 3^* &amp; 11\\ \hline
\end{array}$$
and use the values of $x$ corresponding to the marked points to get the points
$$E=\{ (0,0),(2,\pm 6),(3,\pm 2),(4,\pm 4),(5,0),
(6,\pm 1),(8,0),(9,\pm 6),(11,\pm 4)\}
$$
on the elliptic curve, in addition to the point at infinity, which is the additive identity.