= Solution
For a locally small <category> $\mathcal C$, an object $A$, and a <functor> $F:\mathcal C\to\mathbf{Set}$, the covariant form of the <Yoneda lemma> is the natural <bijection>
$$
\operatorname{Nat}(\mathcal C(A,-),F)\cong F(A),\qquad \alpha\longmapsto\alpha_A(1_A).
$$
For $x\in F(A)$ define $\alpha^x_B(f)=F(f)(x)$, where $f:A\to B$. For $u:B\to D$, functoriality gives $F(u)\alpha^x_B(f)=F(uf)(x)=\alpha^x_D(uf)$, so this is a <natural transformation>. Conversely, naturality of $\alpha$ at $f:A\to B$ gives $\alpha_B(f)=F(f)\alpha_A(1_A)$. The two constructions are inverse. They are natural in $F$, because a <natural transformation> $F\Rightarrow G$ commutes with evaluation. They are also natural in $A$: an arrow $v:A'\to A$ induces $\mathcal C(A,-)\to\mathcal C(A',-)$ by precomposition, and the resulting map $\operatorname{Nat}(\mathcal C(A',-),F)\to\operatorname{Nat}(\mathcal C(A,-),F)$ corresponds to $F(v):F(A')\to F(A)$. Applying the result to the <opposite category> gives $\operatorname{Nat}(\mathcal C(-,A),H)\cong H(A)$ for $H:\mathcal C^{\mathrm{op}}\to\mathbf{Set}$.
Now let $\mathcal C$ be a <small category>, and let $F,G:\mathcal C\to\mathbf{Set}$. Finite <products in a category> in $[\mathcal C,\mathbf{Set}]$ exist pointwise; the terminal <functor> is the constant singleton <functor>. Define the <exponential of covariant set-valued functors> by
$$
(G^F)(C)=\operatorname{Nat}(\mathcal C(C,-)\times F,G).
$$
Smallness ensures that these <natural transformations> form a set. For $u:C\to D$, define $(G^F)(u)(\alpha)$ at $E$ by
$$
\bigl((G^F)(u)(\alpha)\bigr)_E(v,x)=\alpha_E(vu,x),\qquad v:D\to E.
$$
Precomposition respects identities and composition, so this defines a <functor>. Its <evaluation map of an exponential object> is
$$
\mathrm{ev}_C:(G^F)(C)\times F(C)\to G(C),\qquad (\alpha,x)\mapsto\alpha_C(1_C,x).
$$
To check naturality, for $u:C\to D$ both sides give $\alpha_D(u,F(u)x)$: one side uses naturality of $\alpha$, and the other uses the defining action of $G^F$.
Here is the full <universal property>. Given $\theta:H\times F\Rightarrow G$, define $\widehat\theta:H\Rightarrow G^F$ by
$$
\bigl(\widehat\theta_C(h)\bigr)_D(v,x)=\theta_D(H(v)h,x),\qquad v:C\to D.
$$
For $w:D\to E$, naturality of $\theta$ makes this commute with $G(w)$, so $\widehat\theta_C(h)$ is a <natural transformation>. Replacing $h$ by $H(u)h$ is the same as precomposing $v$ by $u$, proving naturality in $C$. Conversely, $\beta:H\Rightarrow G^F$ gives $\theta_C(h,x)=\beta_C(h)_C(1_C,x)$. Starting with $\theta$ recovers it immediately. Starting with $\beta$, its naturality at $v:C\to D$ yields
$$
\beta_D(H(v)h)_D(1_D,x)=\beta_C(h)_D(v,x),
$$
so it too is recovered. The formulas are natural in $H$ and $G$, giving the required <adjunction> $(-)\times F\dashv(-)^F$. Therefore \b[the covariant functor category is cartesian closed]. This proof supplies the actual exponential and does not need a colimit presentation of its arguments.
Finally suppose $\mathcal C$ itself is a <small category> which is a <Cartesian closed category>. In its <presheaf category>, the same construction, applied to $\mathcal C^{\mathrm{op}}$, gives
$$
\begin{aligned}
((YB)^{YA})(C)
&=\operatorname{Nat}(YC\times YA,YB)\\
&\cong\operatorname{Nat}(Y(C\times A),YB)\\
&\cong\mathcal C(C\times A,B)\\
&\cong\mathcal C(C,B^A)\\
&=Y(B^A)(C).
\end{aligned}
$$
The first <isomorphism> uses the product's <universal property>: a map into $C\times A$ is precisely a pair of maps into $C$ and $A$. The next is the <Yoneda lemma>, and the last is the exponential <adjunction> in $\mathcal C$. Each step is natural in $C$, $A$, and $B$. Under these identifications, evaluation corresponds to $Y$ of the evaluation $B^A\times A\to B$. Hence \b[the Yoneda embedding preserves exponentials], with the natural identification $\boxed{Y(B^A)\cong(YB)^{YA}}$.
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