Is the cofinality function $\operatorname{cf}$ monotonic?
I.e., if $\lambda \le \kappa$ for cardinals $\lambda$ and $\kappa$, does it then follow that $\operatorname{cf}(\lambda) \le \operatorname{cf}(\kappa)$?
If this does not hold in general, do there exist conditions on $\lambda$ and $\kappa$ (besides regularity) that allow to make this conclusion in special cases?
No, of course not. If you already know that not all cardinals are regular then it suffices to show that. Simply take $\kappa$ to be a singular cardinal and, $\lambda=\operatorname{cf}(\kappa)^+<\kappa$, then $\lambda$ is regular but $\kappa$ has a strictly smaller cofinality despite being larger.
For example, $\aleph_1<\aleph_\omega$ but $\operatorname{cf}(\aleph_1)=\aleph_1>\aleph_0=\operatorname{cf}(\aleph_\omega)$.
And to your question, no, there are no "easy" cases which are not tantamount to stating "the cofinality function is monotonic in this case".