Find equivalence classes of language $L = \{0^n1^n, n \in \mathrm{N}_0 \}$
Let $\Sigma=\{0,1\}$. Here are all (Myhill–Nerode) equivalence classes, $O_0,O_1, O_2,\cdots, L_0, L_1, L_2, \cdots$ as well as $R$.

$$ \begin{aligned}
O_{k} &amp;= \{0^k\},  \ \text{ where } k \in \mathrm{N_0}. \\
L_{k} &amp;= \{0^{m+k}1^m: m \in \mathrm{N} \},  \ \text{ where } k \in \mathrm{N_0}. \\
R &amp;=\{x10^n1^m: x\in\Sigma^*,\, n,m \in \mathrm{N_0},\, n\ge m\}\cup\{x0^n1^m: x\in\Sigma^*,\, n,m \in \mathrm{N_0},\, n&lt;m\}
\end{aligned} $$

If $a\in O_{k}$ and $b\in O_{t}$, $k\neq t$, $a1^{k}\in L$ while $b1^{k}\not\in L$. So $a$ and $b$ are in different classes.
If $a\in O_{k}$ and $b\in L_{t}$, $a01^{k+1}\in L$ while $b01^{k+1}\not\in L$. So $a$ and $b$ are in different classes.
If $a\in O_{k}$ and $b\in R$, $a1^{k}\in L$ while $b1^{k}\not\in L$. So $a$ and $b$ are in different classes.  

If $a\in L_{k}$ and $b\in L_{t}$, $k\neq t$, $a1^{k}\in L$ while $b1^{k}\not\in L$. So $a$ and $b$ are in different classes.
If $a\in L_k$ and $b\in R$, $a1^k\in L$ while $b1^k\not\in L$. So $a$ and $b$ are in different classes.

What remain to be proved are the following.


For any $k\ge0$, all elements in $O_k$ belong to the same class, which is true trivially since there is only one element. 
For any $k\ge0$, all elements in $L_k$ belong to the same class.
all elements in $R$ belong to the same class.
$\cup_{k=0}^\infty L_k\cup R =\Sigma^*.$


I will let interested readers to fill in the details.



Now that the correct equivalence classes are shown, let me get to your questions.


  In the definition, nothing is said about $x$ and $y$ in $(xR_Ly)$. Do I have to assume that those are two arbitrary words that are already in $L$? (If not, then my approach doesn't work)


No, $x$ and $y$ may or may not be in $L$. Your approach doesn't work.


  For a fixed $n \in \mathrm{N}_0$ I think $\exists$ just one word in the class $C_n$. Then does it mean that the unique word in $C_n$ is only equivalent to itself?


I cannot find where $C_n$ is defined. If it is what I guess it is, there are more than one word in $C_n$. If it means $O_n$ I defined above, then it does have only one word, $0^n$.