3SAT and directed graph Given a 3SAT instance (a Boolean expression in three conjunctural normal form), we draw a directed graph, where for each Boolean variable $x_{i}$ we have the nodes $x_{i}$ and $!x_{i}$; for each clause, for example $\left(x_{a} \vee x_{b} \vee x_{c}\right)$, we draw the following arrows $!x_{a}x_{b}$, $!x_{a}x_{c}$, $!x_{b}x_{a}$, $!x_{b}x_{c}$, $!x_{c}x_{a}$, $!x_{c}x_{b}$.
Is it possible understand from the graph if there is a variable $x_{i}$ such that $x_{i}\Leftrightarrow !x_{i}$ ($!x_{i}$ is $not\left(x_{i}\right)$)?
I think you are trying to build the 2-SAT implication graph for 3-SAT.
In 2-SAT, $(x_a \vee x_b)$ may indeed be considered as 2 implications, $\neg x_a \Rightarrow x_b$ and $\neg x_b \Rightarrow x_a$.
The problem is that $(x_a \vee x_b \vee x_c)$ is not equivalent to any of the 6 implications like $\neg x_b \Rightarrow x_a$, as $x_c$ is sufficient to satisfy the clause.
You may eventually have $(\neg x_a \wedge \neg x_b) \Rightarrow x_c$ or $\neg x_a \Rightarrow (x_b \vee x_c)$, but neither let you build such implication graph.