Denote the diagram:
$\require{AMScd}$ \begin{CD} C_{1} @>{G_{1}}>> C_2 @>{G_{2}}>> C_3\\ @VV{F_1}V @VV{F_2}V @VV{F_3}V\\ D_{1} @>{H_{1}}>> D_2 @>{H_{2}}>> D_3\\ \end{CD}
Where $G_2\circ G_1, H_2\circ H_1,F_2$ are equivalences of categories. I need to prove that $F_1,F_3$ are also equivalences. It's easy to show that $G_1,H_1$ are faithfull, and therfore $F_1$ also, and similarly, $G_2,H_2$ are essentially surjective, and therefore $F_3$. Diagram chasing gives the opposite - $F_1$ is essentially surjective and $F_3$ is faithfull. I couldn't prove that $F_1,F_3$ are full. Any ideas?
First we prove that $F_1$ is full. Let $f: F_1(A) \to F_1(B)$, then $H_1(f): F_2 G_1(A) \to F_2 G_1(B)$. So since $F_2$ is full, we find $f': G_1(A) \to G_1(B)$ with $F_2(f') = H_1(f)$. Now we consider $G_2(f'): G_2 G_1(A) \to G_2 G_1(B)$, and using fullness of $G_2 G_1$ we find $f'': A \to B$ such that $G_2 G_1(f'') = G_2(f')$. Then we chase this construction back: $$ H_2 H_1 F_1(f'') = F_3 G_2 G_1(f'') = F_3 G_2(f') = H_2 F_2(f') = H_2 H_1(f), $$ so $F_1(f'') = f$ by faithfulness of $H_2 H_1$, and we conclude that $F_1$ is full.
To complete the entire argument we can indeed use that $F_3$ is isomorphic to a composition of equivalences (and thus itself an equivalence). We can also use a (not too hard) elementary proof to see that $F_3$ is full.
To see that, let $g: F_3(X) \to F_3(Y)$. Then because $G_2 G_1$ is essentially surjective, there are $X', Y'$ such that $G_2 G_1(X') \cong X$ and $G_2 G_1(Y') \cong Y$. So we have an arrow $$ \bar{g}: F_3 G_2 G_1(X') \xrightarrow{F_3(\cong)} F_3(X) \xrightarrow{g} F_3(Y) \xrightarrow{F_3(\cong)} F_3 G_2 G_1(Y'). $$ Since the isomorphisms are in the image of $F_3$, it suffices to show that $\bar{g}$ is in the image of $F_3$. The domain and codomain of $\bar{g}$ are the same as $H_2 H_1 F_1(X')$ and $H_2 H_1 F_1(Y')$ respectively. So using fullness of $H_2 H_1$ and $F_1$ (as just established), we find $g': X' \to Y'$ such that $$ F_3 G_2 G_1(f') = H_2 H_1 F_1(f') = \bar{g}, $$ and we conclude that $F_3$ is full.