A lattice $L$ is a partially ordered set such that any two elements $a$ and $b$ have a least upper bound $a\lor b$ and a greatest lower bound $a\land b$. A congruence relation on $L$ is an equivalence relation $\sim$ on $L$ that is compatible with the lattice operation in the sense that $a_1\sim b_1$ and $a_2\sim b_2$ implies $a_1\lor a_2\sim b_1\lor b_2$ and similarly for $\land$.
In any lattice there are always two trivial congruence relations, the congruence relation where each element is its own equivalence class (block), and at the other extreme the congruence relation with a single block.
What's an example of a lattice with exactly three congruence relations?
This is Ex 3.40 in Graetzer (Lattice Theory: Foundation). So it is asking to find a lattice with exactly one non-trivial congruence relation. Any example I can think of always has at least two non-trivial congruence relations.
For example, for the linear poset with three elements $a\prec b\prec c$, one congruence relation has $a,b$ in one block and $c$ alone, and another has $a$ alone and $b,c$ together in one block. Another example, the standard diamond lattice with four elements (consisting of all subsets of a two element set ordered by inclusion) also has two non-trivial congruence relations.
Proof: The elements of the diamond right under the top must be either separated or in a single class, because diamond has only trivial congruences.
If they are in a single class, then $d\sim b$ implies $a\sim 0$ and $b\sim e$ implies $c\sim 0$. Thus, $a\sim c$ and $0\sim a\sim a\vee c=1$ and this clearly implies that the congruence is trivial.
Similarly, it is easy to see that $a,b,c$ must be separated, otherwise some pair of elements in the diamond is not separated and we have a trivial congruence again.
So the only possible pairs we can relate are $a\sim d$, $b\sim 0$ $e\sim c$. It is easy to see that any of them forces the other two, so we are left with the single equivalence that is a candidate for a congruence, namely $\{\{0,b\},\{a,d\},\{c,e\},\{f\},\{1\}\}$.
The remaining tedious checking that this is indeed a congruence is omitted.