Let $\mathsf{C},\mathsf{D}$ and $\mathsf{E}$ be categories. Consider a function $\rho\colon\mathsf{Ob}([\mathsf{C\times D},\mathsf{E}])\to\mathsf{Ob}([\mathsf{C},[\mathsf{D},\mathsf{E}]])$ which maps a functor $F\colon\mathsf{C\times D}\to\mathsf{E}$ to the functor $\rho(F)\colon\mathsf{C}\to[\mathsf{D},\mathsf{E}]$ such that
for any $X \in \mathsf{C}$, $\rho(F)(X)$ is the functor which sends each $A \in \mathsf{D}$ to $F(X,A)$ and each morphism $g\colon A\to B$ to $F(1_X,g)$,
for any morphism $f\colon X\to Y$ in $\mathsf{C}$, $\rho(F)(f)$ is the natural transformation $\rho(F)(X)\Rightarrow\rho(F)(Y)$ so that for any $A \in \mathsf{D}$ we have $\rho(F)(f) = F(f,1_A)$.
I'm trying to prove that the function is surjective. Here's what I did so far:
Let $H\colon\mathsf{C}\to[\mathsf{D},\mathsf{E}]$ be a functor. Define a hypothetical functor $F\colon\mathsf{C\times D}\to\mathsf{E}$ which sends each pair $(X,A)$ to $H(X)(A)$ and each pair $f\colon X\to Y$ and $g\colon A\to B$ of morphisms to $H(f)_B\circ H(X)(g)$. It's not hard to show that this construction preserves domains, codomains and identites.
But I don't know how to show that such $F$ preserves composition between morphism. Perhaps, we have to use naturality somehow.
Let $X,Y,Z\in\mathsf{Ob}(\mathsf{C})$ and $A,B,C\in\mathsf{Ob}(\mathsf{D})$, $f\colon X\to Y$, $g\colon A\to B$, $h\colon Y\to Z$, $k\colon B\to C$. Then: $$ F((h,k)\circ(f,g))=F(h\circ f,k\circ g)=(H(h\circ f))(C)\circ(H(X))(k\circ g)= $$ $$ =(H(h))(C)\circ(H(f))(C)\circ(H(X))(k)\circ(H(X))(g)= $$ $$ =(H(h))(C)\circ(H(Y))(k)\circ(H(f))(B)\circ(H(X))(g)= $$ $$ =(H(Z))(k)\circ(H(h))(B)\circ(H(Y))(g)\circ(H(f))(A)= F(h,k)\circ F(f,g) $$ by functoriality and naturality. Actually (as you are probably aware of) the functor $\rho\colon[\mathsf{C}\times\mathsf{D},\mathsf{E}]\to[\mathsf{C},[\mathsf{D},\mathsf{E}]]$ is an isomorphism of categories, so the mapping $\rho_{\mathsf{Ob}}\colon\mathsf{Ob}([\mathsf{C}\times\mathsf{D},\mathsf{E}])\to\mathsf{Ob}([\mathsf{C},[\mathsf{D},\mathsf{E}]])$ is a bijection.