Let $\mathcal C \to \mathcal D \to \mathcal E$ be a pair of functors. Assume that $\mathcal C$ is dense in $\mathcal E$. Is then $\mathcal D$ also dense in $\mathcal E$? One definition of denseness is that the nerve, that is the restricted Yoneda embedding $\mathcal E \to \mathrm{PSh}(\mathcal C)$ is fully faithful. We want that the same holds with $\mathcal D$ instead of $\mathcal C$. Everything fits into a commutative diagram
$$\require{AMScd} \begin{CD} \mathcal E @>>> \mathrm{PSh}(\mathcal D)\\ @. {_{}\style{display: inline-block; transform: rotate(30deg)}{{\xrightarrow[\rule{4em}{0em}]{}}}} @VVV\\ @. \mathrm{PSh}(\mathcal C), \end{CD}$$
where the vertical functor is the restriction. Obviously the diagonal functor is faithful because the top is faithful but from there I would need something like the restriction being faithful for example, which is not always the case. So is there a counterexample to the question or some (reference to a) proof?
A useful reference for cancellability of dense functors is §5.2 of Kelly's Basic Concepts of Enriched Category Theory. For instance, in your example, if $\mathcal C \to \mathcal D$ is essentially surjective, or if $\mathcal D \to \mathcal E$ is fully faithful, then density of $\mathcal C \to \mathcal D \to \mathcal E$ implies density of $\mathcal D \to \mathcal E$ (Proposition 5.11 and Theorem 5.13 ibid.).