Functors Preserve Isomorphisms

Since $f$ is an isomorphism, there exists a morphism $$ f^{-1} : Y \to X $$ such that $$ f^{-1} \circ f = \operatorname{id}_X, \qquad f \circ f^{-1} = \operatorname{id}_Y . $$ Apply the functor $F$ to the first equality: $$ F(f^{-1} \circ f) = F(\operatorname{id}_X). $$ By functoriality of composition and identities this becomes $$ F(f^{-1}) \circ F(f) = \operatorname{id}_{F(X)}. $$ Applying $F$ to the second equality gives $$ F(f) \circ F(f^{-1}) = \operatorname{id}_{F(Y)}. $$ Thus $F(f^{-1})$ is an inverse of $F(f)$, and therefore $F(f)$ is an isomorphism.