Does it help for clique if the vertices are partitioned into 3 cliques?
Computing a maximum clique in a graph whose vertex set can be partitioned into 3 cliques, is equivalent to computing a maximum independent set in the complement graph, which can be partitioned into 3 independent sets and is therefore 3-colorable. The latter problem is still NP-complete, which can be seen as follows: Start from an arbitrary instance $(G,k)$ of independent set and subdivide each edge twice, i.e., replace each edge ${u,v}$ by a path ${u,e_1, e_2, v}$ on two new vertices. It is not difficult to prove that the resulting graph $G'$ has an independent set of size $k + |E(G)|$ if and only if $G$ has a size-$k$ independent set: for every independent set in $G$, you can add exactly one of the two subdivider vertices on every edge without violating independence.

Hence the Clique problem is still NP-complete for such inputs.