I have been trying to think of a way to characterize the hom functors via universal property. I could not find any such thing elsewhere online. So I came up with a property, inspired by Yoneda lemma, and a proof that any functor satisfying this property must be representable:
For a category $C$, let $R\in Set^{C^{op}}$ such that $Hom_{Set^{C^{op}}}(R,F)\cong Fc$, natural in $F$. Then $R\cong C(-,c)$. This would lend the universal arrow $(R, r : 1 \to Rc)$ from $1=\{0\}$ to the functor that sends $F\mapsto Fc$ and $(\alpha: F\Rightarrow G)\mapsto (\alpha)_c$. The element $r\in Rc$ is defined in the proof, sorry for putting that before hand. But it is basically the analog of $1_c\in C(c,c)$.
I will include a quick proof that any such $R$ satisfies $R\cong C(-,c)$. But, before that, here is my question: is there something like this already out there, i.e., a characterization of representable functors via universal properties? Does this suffice to define hom functors? Of course this will not give us the hom functor on the nose, but it lends something isomorphic. Since category theory is all upto isomorphism, shouldn't this suffice? Or are there draw backs to taking this as the definition of a representable functor?
The proof sketch is roughly as follows:
For any $F$ and any $x\in Fc$, let $\Psi_x$ denote the natural transformation $R\Rightarrow F$ given by the hypothesis. Now observe that $Hom(R,R)\cong Rc$ lends $r\in Rc$, determined by $1_R:R\Rightarrow R$.
Lemma: for any $F$ and any $x\in Fc$, $(\Psi_x)_c(r)=x$, and $\Psi_x$ is the unique such $\Psi$. The naturality of $F$ implies the following diagram commutes: 
Chasing $1_R$ through lends the desired equality. Uniqueness follows from the isomorphism.
Now by the Yoneda lemma, $r$ determines $\Phi_r : C(-,c)\Rightarrow R$. Further, the assumption quickly lend that $1_c : c \to c$ in $C(c,c)$ determines $\Psi_{1_c} :R\Rightarrow C(-,c)$.
Now, observe that $(\Phi_r\cdot\Psi_{1_c})_c(r)=(\Phi_r)_c(1_c)=r$, applying the lemma and then Yoneda. But, by the uniqueness claim of the lemma, we have $(\Phi_r\cdot\Psi_{1_c})=1_R$. In the other direction, $(\Psi_{1_c}\cdot\Phi_r)_c(1_c)=(\Psi_{1_c})_c(r)=1_c$. By Yoneda and then the lemma. By Yoneda, it follows that $(\Psi_{1_c}\cdot\Phi_r)=1_{C(-,c)}$. Thus, $\Psi_{1_c}:R\cong C(-,c)$. QED
Maybe this is all just a trivial consequence of the Yoneda lemma, but I have never seen hom functors, nor representable functors presented this way.
Best Answer
Here is a way to use the Yoneda lemma to quickly prove the statement you're interested in: by the Yoneda lemma, we also have $\hom(\hom(-,c),F)\cong Fc$, naturally in $F$.
Therefore $\hom(\hom(-,c),F)\cong \hom(R,F)$ naturally in $F$. By the Yoneda lemma, applied to $Fun(C^{op},Set)$, $\hom(-,c)\cong R$.
Note that representable functors come with the given isomorphism $\hom(R,F)\cong Fc$. If you have this, then $R$ is just as good as $\hom(-,c)$, indeed, and this is equivalent to being representable. But it's not a definition, just an equivalent characterization.
The obvious drawback is that the word "representable" becomes less close to the definition if you define it this way. But it is definitely widely used, yes !