Proving a language is neither Recursively Enumerable nor co-Recursively Enumerable
One way to solve this problem is by showing there do not exist Turing machines that accept  $L_3$ or $\overline L$. But probably that is simply restating the question. 

$L_u$, the universal language, is non-recursive but RE language. If we reduce $L_u \leq_M L$ then $L$ cannot be co-RE. If we reduce $L_u \leq_M \overline{L}$ then $L$ cannot be RE.

Let us  first do the reduction $L_u \leq_M L$.

We shall reduce $\langle M,w\rangle$ to a Turing machine $M'$ as follows:
$M'$ rejects all inputs other than $w$. If the input is $w$ then it runs as Turing Machine $M$ on $w$. This gives us $\langle M,w\rangle \in L_u$ iff $M' = f(\langle M,w\rangle) \in L$.

Hence $L_u \leq_M L$. Therefore $L$ cannot be co-RE. Otherwise $L_u$ will become recursive.

Next we do the reduction $L_u \leq_M \overline L$.

We shall reduce $\langle M,w\rangle$ to a Turing machine $M''$ as follows: Let $w'$ be a string other than $w$.
$M''$ rejects all inputs other than $w$ and $w'$. If the input in $w'$ then $M''$ accepts the input. If the input is $w$ then it runs as Turing Machine $M$ on $w$. This gives us $\langle M,w\rangle \in L_u$ iff $M''= f(\langle M,w\rangle) \not\in L$, that is, $\langle M,w\rangle \in L_u$ iff $M'' \in \overline L$.

Hence $L_u \leq_M \overline{L}$. Therefore $L$ cannot be RE. Otherwise $L_u$ will become recursive.

If we combine the results, $L$ is neither RE nor co-RE.