Here is an interesting construction that may be relevant, though it doesn't quite answer the question.

Firstly, note that as Theo says, it suffices to assume $X=1$, i.e. to ask whether $C(1,Z^Y) \cong C(Y,Z)$. For if this is true, then for any $X$ we have

$$ C(X\times Y,Z) \cong C(1, Z^{X\times Y}) \cong C(1,(Z^Y)^X) \cong C(X,Z^Y). $$

Secondly, as Theo also says, note that by adjointness and the Yoneda lemma, it suffices to ask whether $(A\odot 1)\times Y\cong A\odot Y$ for all $A\in V$, where $\odot$ denotes the copower of $C$ over $V$. For we have isomorphisms

$$ V_0(A,C(1,Z^Y)) \cong C_0(A\odot 1, Z^Y) \cong C_0((A\odot 1)\times Y,Z) $$

$$ V_0(A,C(Y,Z)) \cong C_0(A\odot Y,Z).$$

Now this question $(A\odot 1)\times Y\cong A\odot Y$ does not refer to cartesian closedness of $C$. And if we drop the hypothesis that $C$ is cartesian closed, then this isomorphism can fail, even when $C$ is complete and cocomplete and otherwise nice (even locally presentable, if $V$ is). To show this, it suffices to exhibit a *small* $V$-category $D$ containing an object $Y$, a terminal object $1$, copowers $A\odot 1$ and $A\odot Y$, and a product $(A\odot 1)\times Y$, for some $A\in V$, such that $(A\odot 1)\times Y\ncong A\odot Y$. For if we have such a $D$, let $C$ be the full subcategory of the functor $V$-category $[D^{\rm op},V]$ spanned by those functors $F$ that preserve the two copowers $A\odot 1$ and $A\odot Y$. Then $C$ is reflective in $[D^{\rm op},V]$, hence complete and cocomplete (and locally presentable, if $V$ is), and the embedding $D\hookrightarrow C$ preserves the two copowers, as well as the terminal object and the product, and is fully faithful so it also preserves the non-isomorphism. (But, there seems no reason in general why $C$ should be cartesian closed.)

Finally, to produce such a $D$ we can use the time-honored technique of looking at a universal example. Consider the $V$-category $D$ with *exactly* those five objects, which is "freely generated" by the requirement that the two copowers, the terminal object, and the product have their appropriate universal properties. The hom-objects of this category are given in the following table ($D(x,y)$ is in the $x$-row and the $y$-column).

$$
\begin{array}{c|ccccc}
\nearrow & A\odot 1 & A\odot Y & Y & (A\odot 1)\times Y & 1 \\\hline
A\odot 1 & A^A & \emptyset & \emptyset & \emptyset & 1 \\
A\odot Y & A^A & A^A & 1 & A^A & 1 \\
Y & A & A & 1 & A & 1 \\
(A\odot 1)\times Y & A^A & A & 1 & A^A & 1 \\
1 & A & \emptyset & \emptyset & \emptyset & 1 \\
\end{array}
$$

(This is assuming there are no morphisms $A\to \emptyset$ in $V$, as is the case whenever $A$ is not initial since $V$ is cartesian closed.) I will leave it as an exercise for the reader to define all 125 composition maps, check the 50 identity equations and 625 associativity equations to be sure we have a category, and check that 20 induced maps are isomorphisms to be sure we have the right universal properties (we don't need to check "freeness" of $D$; all that matters is that it exists, although we used the idea of freeness to construct it). And before you ask, yes, I actually did that exercise myself, although Coq helped a lot. (-:

In particular, the composition map

$$D((A\odot 1)\times Y, A\odot Y) \times D(A\odot Y, (A\odot 1)\times Y) \to D(A\odot Y,A\odot Y)$$

is the morphism $A \times A^A \to A^A$ whose adjunct $A\times A^A \times A\to A$ is projection onto the first copy of $A$. In other words, $A \times A^A \to A^A$ factors through the "inclusion of constant functions" $A\to A^A$, and thus as long as $A$ is nontrivial it can never yield the identity. Hence $(A\odot 1)\times Y\ncong A\odot Y$ in $D$.

So to summarize: the question can be stated in an equivalent form about copowers, and if we drop the assumption of cartesian closedness then the answer to that related question becomes false. So if the statement is true, it would have to be some kind of exactness property following from cartesian closedness, not due to any general fact about copowers.

1more comment