Prove that if $L$ is regular then $f(L)$ is regular Prove that if $L$ is regular then $f(L)$ is regular.
$$f(L)=\{w: \text{every prefix of $w$ of odd length $\in$ L } \}$$
So my attemption is:
Let $M= (Q, \Sigma, \delta, q_0,F) $ will be DFA recognizing $L$
Let construct $M'=(Q', \Sigma,\delta', q_0', F')$ recognizing $f(L)$.
$Q'=Q\times\{1,2\}$
$F'=\{(q,1)|q\in F\}\cup\{(q,2)|q\in F\}$
$q_0'= (q_0, 1)$
$\delta'((q,1),a\in\Sigma)=\begin{cases} (\delta(q,a),2);\delta(q,a)\in F\\ \emptyset;  \delta(q,a)\notin F\end{cases}  $
$\delta'((q,2), a\in\Sigma)=(\delta(q,a),1)$  
What about this solution ?

You can prove the result without using automata. Let $A$ be the alphabet, and let $$
K = \{ u \in A^* \mid \text{ every prefix of $u$ of odd length is in $L$} \}
$$
Let $K^c$ be the complement of $K$ in $A^*$. Then
\begin{align}
K^c &= \{ u \in A^* \mid \text{ there exists a prefix of $u$ of odd length in $L^c$} \}\\
&= \{ u \in A^* \mid \text{ there exists a prefix of $u$ in $(A^2)^*A \cap L^c$} \}\\
&= \bigl((A^2)^*A \cap L^c\bigr)A^*
\end{align}
Therefore $K = \Bigl(\bigl((A^2)^*A \cap L^c\bigr)A^*\Bigr)^c$ and since regular languages are closed under intersection, complement, product and star, $K$ is regular.
