I only have proved that if graf($f$) is recursive set then is recursively enumerable set by $(n_0, m_0) \in Graf(f) $ $$h(n):= \begin{cases} (n, m) \textit{ If } (n,m) \in Graf(f) \\(n_0, m_0) \textit{ If no} \end{cases}$$ which is recursive.
I know that $Graf(f)$ is a recursive set iff the characteristic $$Graf(f)(n,m):= \begin{cases} 1 \textit{ if } (n,m) \in Graf(f) \\0 \textit{ If no} \end{cases}$$ is a recursive function.
Edit: From the definition if $Graf(f)$ is recursively enumerable then there exists a recursive predicate $R$ such that $$(x, y) \in Gra(f) \Leftrightarrow \exists z R(x, y, z)$$ Thus, $f(x)= (\mu z [R(x, (z)_0, (z)_1])_0$ is recursive, where $()_i$ is the recursive function of Gödel for tuples.
Furthermore, if $f$ is a recursive function then $$(x, y) \in Graf(f) \Leftrightarrow f(x)=y$$ is a recursive set because $=$ is a recursive relation.