We are to show that in a Cartesian closed category (ccc) $\mathcal{C}$, that for any objects $X, Y, Z$, there exists a natural isomorphism $(Y \times Z)^X \cong Y^X \times Z^X$.
Disclaimer: all of this is in preparation of, so 'not using' the Yoneda Lemma (or adjunctions). It should be doable by the the definition of the exponential itself (and of course the relevant universal property of the product).
First off, I don't think (nor hope) the use of 'natural' here is the one that is defined earlier in the notes (being a natural transformation that is an isomorphism in, in this case, the endofunctor category $[\mathcal{C}, \mathcal{C}]$, i.e., an isomorphism at each component. This would require the objects to be lifted to functors and things will quickly get messy and tedious..
So I will suppose the use of 'natural' here is colloquial. In my text, being an exponential of $Y\times Z$ and $X$ means that there is a pair of an object and a morphism $((Y \times Z)^X, \text{ev} : (Y \times Z)^X \times X \to Y \times Z)$, having a certain universal property that can be expressed as a diagram but also as the fact that for each object $A$,
\begin{align*} \text{Hom}_{\mathcal{C}}(A, (Y \times Z)^X) &\to \text{Hom}_{\mathcal{C}}(A \times X, Y \times Z) \\ H &\mapsto \text{ev} \circ (H \times \text{id}_{Y \times Z}) \end{align*}
is a bijection.
Is the idea to use the map given by the universal property of th product $Y^X \times Z^X$, or is there some other trick of matching/gluing the diagrams for products and exponentials together, or is it easier to just work with Hom-sets here? Or something else entirely?
EDIT
I actually do prefer to proceed with the Hom-sets. Namely (I write $\leftrightarrow$ for bijection), by the product properties and exponential properties, respectively, we have, for all objects $A$, bijections
\begin{align*} \text{Hom}_{\mathcal{C}}(A, Y^X \times Z^X) &\leftrightarrow \text{Hom}_{\mathcal{C}}(A, Y^X) \times \text{Hom}_{\mathcal{C}}(A, Z^X) \\ &\leftrightarrow \text{Hom}_{\mathcal{C}}(A \times X, Y) \times \text{Hom}_{\mathcal{C}}(A \times X, Z). \end{align*}
But then, using the universal property of the exponential again, we obtain
\begin{align*} \text{Hom}_{\mathcal{C}}(A, (Y \times Z)^X) &\leftrightarrow \text{Hom}_{\mathcal{C}}(A \times X, Y \times Z) \\ &\leftrightarrow \text{Hom}_{\mathcal{C}}(A \times X, Y) \times \text{Hom}_{\mathcal{C}}(A \times X, Z) \\ &\leftrightarrow \text{Hom}_{\mathcal{C}}(A, Y^X \times Z^X). \end{align*}
Now I am beginning to see how powerful Yoneda is, I think. (I do know the theorem, and therefore I strongly believe that this is it, as we have now derived $\text{Hom}_{\mathcal{C}}(A, Y^X \times Z^X) \leftrightarrow \text{Hom}_{\mathcal{C}}(A, (Y\times Z)^X)$. Since this holds for all $A$, I would presume that this translates to the functor isomorphism $\text{Hom}_{\mathcal{C}}(-, Y^X \times Z^X) \cong \text{Hom}_{\mathcal{C}}(-, (Y\times Z)^X)$.
I think I am therefore there, but if anyone could close the loop on this would be even better :)
Think I've closed the loop myself: since the bijections $\phi_A : \text{Hom}_{\mathcal{C}}(A, (Y\times Z)^X) \leftrightarrow \text{Hom}_{\mathcal{C}}(A, Y^X \times Z^X)$ exists for all objects $A$ (in particular for $A := Y^X \times Z^X$), and since $\text{id}_{Y^X \times Z^X} \in \text{Hom}_{\mathcal{C}}(Y^X \times Z^X, Y^X \times Z^X)$, surjectivity of the corresponding $\phi_{Y^X \times Z^X}$ now ensures that this identity is reached. Thus, for some morphism $f : (Y\times Z)^X \to Y^X \times Z^X$, we have
$$\text{id}_{Y^X \times Z^X} = f\circ \phi_{Y^X \times Z^X}(f),$$
obtaining a right-inverse of $f$.
By the injectivity of the $\phi_A$, this morhpism is unique with this property.
Now we can choose $A:= (Y \times Z)^X$ and apply the argument in reverse (using $\phi_{(Y \times Z)^X}^{-1}$) to find a (unique) left-inverse of $f$, thereby obtaining a unique isomorphism $(Y \times Z)^X \to Y^X \times Z^X$.
Being unique, this isomorphism should be pretty natural, I would presume ^^