$\cong\;=\{((x_1,y_1), (x_2,y_2))\in \mathbb R^2 ×\mathbb R^2 |x_1^2-x_2^2=3y_1^2-3y_2^2\}$ finitary relation meaning $(x_1,y_1) \cong (x_2,y_2)$ if $x_1^2-x_2^2=3y_1^2-3y_2^2$ Is this finitary relation reflexive, symmetric, antisymmetric, transitive? Is it linear order? Is it equivalence?
Reflexivity: yes (?) since for example (2,2), (2,2): $2^2-2^2=3*2^2-3*2^2$ Symmetry: no (?) since for example (2,5), (5,2): $2^2-5^2\neq3*5^2-3*2^2$ Did I get it correctly? I'm not sure if I got the definitions of reflexivity, symmetry, etc. right...
I think you are confusing reflexivity and symmetry in the relation with those properties within each ordered pair. For example, if you want to find a counter-example for the symmetry property, you need to show $((a,b),\;(c,d)) \in\; \cong $ but $((c,d),\;(a,b)) \not\in\; \cong$ for some $a,b,c,d \in \mathbb{R}$.
Also, if you want to prove a property is true, such as for reflexivity, you can't show it's true just for a particular example. You have to show it's true for all values.
Reflexive? Yes. For all $(x,y) \in \mathbb{R}^2,\; x^2 - x^2 = 0 = 3y^2 - 3y^2 \quad$ so $((x,y),\;(x,y)) \in \;\cong$.
Symmetric? Yes. If $((x_1, y_1),\; (x_2,y_2)) \in \;\cong\;$ then $$x_2^2 - x_1^2 = -(x_1^2 - x_2^2) = -(3y_1^2 - 3y_2^2) = 3y_2^2 - 3y_1^2.$$
So $((x_2, y_2),\; (x_1,y_1)) \in \;\cong$.
Antisymmetric? No. Counter-example:
$((0,1),\;(0,-1))$ and $((0,-1),\;(0,1))$ are in $\cong $ but $(0,1) \neq (0,-1)$.
Transitive? Yes. Suppose $((x_1, y_1),\; (x_2,y_2)) \in \;\cong \;$ and $((x_2, y_2),\; (x_3,y_3)) \in \;\cong$. Then,
\begin{eqnarray*} x_1^2 - x_3^2 &=& (x_1^2 - x_2^2) + (x_2^2 - x_3^2) \\ &=& (3y_1^2 - 3y_2^2) + (3y_2^2 - 3y_3^2) \\ &=& 3y_1^2 - 3y_3^2. \end{eqnarray*}
So $((x_1, y_1),\; (x_3,y_3)) \in \;\cong$.
A Linear Order? No, because $\cong\;$ is not antisymmetric.
An Equivalence Relation? Yes, because $\cong\;$ is reflexive, symmetric and transitive.