If two categories are equivalent, then if one has products, then so does the other. The proof of this is easy enough so I'm guessing the same result holds for exponentials but I am having trouble proving it. Here is what I have so far:
If $F:C\Leftrightarrow D:G$ denotes the equivalence between categories $C$ and $D$, and $f:U$x $V\rightarrow W$, we seek an object $W^{V}$, satisfying the definition of exponential. I know that if $U$ x $V$ is a product in $C$, then it is isomorphic to $G(FU$ x $ FV)$. Applying $F$ and using the natural isomorphism $α:FG\rightarrow 1_{D}$, we get $F(U$ x $ V)≅FU$ x $FV$.Thus we have the arrow $Ff:FU$ x $FV\rightarrow FW$, and the exponential $FW^{FV}$ in $D$.
I want to prove $G(FW^{FV})$ is an exponential object $W^{V}$ in $C$:
We have unique morphism $\tilde{Ff}:FU\rightarrow FW^{FV}$ and the morphisms $\tilde{Ff}$ x $1_{FV}:FU$ x $FV\rightarrow FW^{FV}$ x $FV$ and $ε:FW^{FV}$ x $FV→FW$ s.t. $\varepsilon ∘(\tilde{Ff}$ x $1_{FV})=Ff$.
Now apply $G$ to get $G\varepsilon ∘G(\tilde{Ff}$ x $1_{FV})=GFf. $We have that $GFf$ is an arrow from $G(FU$ x $FV)$ to $GFW$, the domain and codomain of which are isomorphc to $U$ x $V$ and $W$, respectively, so we get $f:U$ x $V\rightarrow W$ back via these isomorphisms. But I do not see how to handle $G(\tilde{Ff}$ x $1_{FV})$, which I want to say gives us the required $\tilde{f}$ x $1_{V}$.
The Yoneda lemma is a very powerful tool, use it. The definition of an exponential object in a category $C$ is (where this is a natural isomorphism in $U$): $$\hom_C(U \times V, W) \cong \hom_C(U, W^V)$$
So here (I suggest you carefully check for each step which property I used): $$\begin{align} \hom_C(U \times V, W) & \cong \hom_C(U \times V, GF(W)) \\ & \cong \hom_D(F(U \times V), F(W)) \\ & \cong \hom_D(F(U) \times F(V), F(W)) \\ & \cong \hom_D(F(U), F(W)^{F(V)}) \\ & \cong \hom_C \left( U, G \left( F(W)^{F(V)} \right) \right) \end{align}$$
Each of these isomorphisms are natural, therefore $G \left( F(W)^{F(V)} \right)$ is an exponential object $W^V$.