(i) looks good.
(ii) doesn't look so good. You didn't prove the general case, you only showed it for $\lambda = 1$. You want it to be true for any nonzero real value of $\lambda$. You should have if $(x_1,y_1) \sim (x_2,y_2)$, then by definition $(x_1,y_1)=(\lambda x_2,\lambda y_2)$, so then let $\gamma = \frac{1}{\lambda}$, and then we have $(x_2,y_2) = (\gamma x_1,\gamma y_1)$, so $(x_2,y_2) \sim (x_1,y_1)$.
(iii). Again, you didn't prove the general case. You should have: if $(x_1,y_1) \sim (x_2,y_2)$, then $(x_1,y_1) = (\lambda x_2, y_2)$. If $(x_2,y_2) \sim (x_3,y_3)$, then we have, where $\alpha$ is some nonzero real, $(x_2,y_2) = (\alpha x_3, \alpha y_3)$. So then $(x_1,y_1) = (\lambda \alpha x_3, \lambda \alpha y_3)$. The reals are closed under multiplication so $\lambda\alpha$ is a real, so it works.
An equivalence class is just a set of elements that are equivalent under the equivalence relation. For example, from (iii), all $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ would be in an equivalence class, because they are all equivalent. If there are any other $(x_n,y_n)$ that are equivalent to them, then they would also be in that class.
You have correctly wrote a proof of the first two properties, reflexivity and symmetry. For transitivity, you did not quite write things as you should. Don't be afraid to explicitly write the elements of $S$! I wrote a short proof below so you can see what I mean:
Denote $\mathbb{N}$ the set of positive integers. Like @Anurag A mentioned here, for ordered pairs $(x_1,x_2), (y_1,y_2) \in \mathbb{N} \times \mathbb{N} $, the definition of $(x_1,x_2)R(y_1,y_2)$ can be restated as
$$x_1 - x_2 = y_1 - y_2.$$
With this in mind, let's prove:
Reflexivity: for all $(x_1,x_2) \in \mathbb{N} \times \mathbb{N} $, of course we have that $x_1 - x_2 = x_1 - x_2$;
Symmetry: for all $(x_1,x_2), (y_1,y_2) \in \mathbb{N} \times \mathbb{N} $, if $x_1 - x_2 = y_1 - y_2$, then exchanging the left side with the right one (this is just the symmetry of the "$=$" relation), we check that $y_1 - y_2 = x_1 - x_2$;
Transitivity: for all $(x_1,x_2), (y_1,y_2),(z_1,z_2) \in \mathbb{N} \times \mathbb{N} $, if $x_1 - x_2 = y_1 - y_2$ and $y_1 - y_2 = z_1 - z_2$, then just equal both equations and we get
$$x_1 - x_2 = z_1 - z_2,$$
so $(x_1, x_2) R (z_1,z_2)$.
For the number of equivalence classes, as Anurag A said, two pairs in $\mathbb{N} \times \mathbb{N}$ are equivalent if the difference of their first and second coordinates is the same. So you have an equivalence class for each possible difference: $\ldots, -2,-1,0,1,2, \ldots$, that is, for every integer. The cardinality of the set of equivalence classes is the same as $\mathbb{Z}$.
Finally, as a piece of advice, I understand that it is helpful for you to write the definitions of things you want to prove and then proceed to prove them (often, ending up writing the exact same thing twice). I recommend you do this on your notes of course, but when showing your work to other people (like here or to your teacher), you can be more clear if you just write directly what you are going to prove. The reader should know what the defintions involved are, anyway.
Best Answer
Tol prove it's reflexive, let $(x, y)$ be any point of $R^2$,. You want to show that $(x, y) \sim (x, y)$. Write that out: it says that
$$x^2 + y^2 = x^2 + y^2$$
which is evidently true. So you've proved reflexivity. Now...you give symmetry a try,...