Let $\mathcal{C}$ be a cartesian closed category. If we are given a morphism $f: Y \to Z$ and some object $X$ in $\mathcal{C}$, is there a nice way to obtain a morphism $Y^X \to Z^X$?
In Set this is true because we can take $f \circ -$, but I have not found a way to generalize this to arbitrary ccc's.
Yes.
By the universal property of exponential objects, morphisms $Y^X \to Z^X$ correspond with morphisms $Y^X \times X \to Z$, so take the morphism $f^X : Y^X \to Z^X$ corresponding with $$Y^X \times X \xrightarrow{\mathrm{ev}} Y \xrightarrow{f} Z$$ where $\mathrm{ev}$ is the evaluation morphism.
A similar trick shows that a morphism $f : Y \to Z$ gives rise to a morphism $X^f : X^Z \to X^Y$, namely the one corresponding with $$X^Z \times Y \xrightarrow{\mathrm{id} \times f} X^Z \times Z \xrightarrow{\mathrm{ev}} X$$
You can prove that the assignments $f \mapsto f^X$ and $f \mapsto X^f$ are functorial (covariantly and contravariantly, respectively).