Emily Riehl's "Category Theory in Context, ${\rm Exercise}~2.1.{\rm iii}.$
Suppose $F:{\rm C}\to{\rm Set}$ is equivalent to $G:{\rm D}\to{\rm Set}$ in the sense that there is an equivalence of categories $H:{\rm C}\to{\rm D}$ so that $GH$ and $F$ are naturally isomorphic.
$~~~~~~(i)$ If $G$ is representable, then $F$ is representable?
$~~~~~(ii)$ If $F$ is representable, then $G$ is representable?
Proof of $(i)$:
Consider the following diagram
$$\require{AMScd}\begin{CD}
Fc @>{\alpha_c}>> GHc @>{\mu_c}>>{\rm Hom}(Hc^*,Hc) @>{\eta_c}>>{\rm Hom}(c^*,c)\\
@V{Ff}VV @V{GHf}VV @V{Hf\circ-}VV @V{f\circ-}VV\\
Fc' @>{\alpha_{c'}}>> GHc' @>{\mu_{c'}}>>{\rm Hom}(Hc^*,Hc') @>{\eta_{c'}}>>{\rm Hom}(c^*,c')
\end{CD}$$
The first square commutes by the naturality of $\alpha:F\Rightarrow GH$, the second square commutes as $G$ is representable and thus $G\cong{\rm Hom}(d^*,-)$ for some $d^*\in{\rm D}$ (decoded by the maps $\mu_c,~\mu_{c'}$). As $H$ is an equivalence of categories it is in particular essentially surjective on objects and thus $d^*$ is isomorphic to the image of an element $c^*\in{\rm C}$, that is $d^*\cong Hc^*$. Furthermore, the equivalence $H$ asserts that there is a bijection $\eta$ between the ${\rm Hom}$-sets of ${\rm C}$ and ${\rm D}$ for any objects $c,~c'$ (such that $f:c\to c'$) guaranteeing the commutivity of the last square.
Since every intermediate square commutes, we have a commutative rectangle from which it follows that $F\cong{\rm Hom}(c^*,-)$, so $F$ is representable.
Proof of $(ii):$
Consider the following diagram
$$\require{AMScd}\begin{CD}
GHc @>{\alpha'_c}>> Fc @>{\mu'_c}>>{\rm Hom}(c^*,c) @>{\eta'_c}>>{\rm Hom}(Hc^*,Hc)\\
@V{GHf}VV @V{Ff}VV @V{f\circ-}VV @V{Hf\circ-}VV\\
GHc' @>{\alpha_{c'}}>> Fc' @>{\mu'_{c'}}>>{\rm Hom}(c^*,c') @>{\eta'_{c'}}>>{\rm Hom}(Hc^*,Hc')
\end{CD}$$
In a similiar manner to $(i)$, the first square commutes by the naturality of $\alpha:GH\to F$ (using the opposite direction given due to the isomorphism), the second square commutes as $F$ is representable by some object $c^*\in{\rm C}$, and the last square commutes by the bijection of ${\rm Hom}$-sets induced by the equivalence $H$ (again, utilizing the opposite direction).
Since every intermediate square commutes, we have a commutative rectangle from which it follows that $G\cong{\rm Hom}(Hc^*,-)$, so $G$ is representable.$~~~\square$
Is my proof correct? If so, can it be improved; if not, where did I went wrong?
Thanks in advance!
Your proof is essentially correct, you could improve (of course, whether or not it's an improvement depends on who you ask) it by making it into a series of reduction steps in the following way :
(a) If $F\cong K$ and $K$ is representable, so is $F$. Proof : trivial, as natural isomorphism is transitive.
(a) reduces the question to proving that $GH$ is representable. But also :
(b) If $G\cong K$ then $GH\cong KH$. Proof : easy, as if $\eta$ is a natural iso, so is $\eta H$.
(b) reduces the question to $\hom (d,-)\circ H$. At this point you have a smaller diagram to draw (the little square at the right of your diagram), so it's a bit easier.
And your (ii) can be improved even more (depending on your definition of equivalence - if you take it to mean "there is a quasi-inverse"). Indeed, if there is a quasi-inverse $L$ to $H$, then you can just apply your (i) to $G$ and $FL$, given that $FL \cong GHL \cong G$ (use (b) at some point)