Consider a commutative diagram of complexes over a ring $R$ $$\require{AMScd}\begin{CD} 0 @>>> X @>>> Y @>>> Z \\ @. @VVfV @VVgV @VVhV \\ 0 @>>> X' @>>> Y' @>>> Z' \end{CD}$$ with exact rows. I know that the complexes $Y$, $Y'$, $Z$, and $Z'$ are exact, hence the maps $g$ and $h$ are automatically quasi-isomorphisms. (i.e., by definition, they induce isomorphisms on the homologies).

My question is: Is the leftmost vertical map $f$ also a quasi isomorphism?

A natural attempt is to use the long exact sequences of homology modules induced by short exact sequence of complexes. However, I'm not able to conclude it.