Let we have a regular graph. I want to know if we can partition the vertex set of this graph while in any part there exist a vertex with all its neighborhood?

You are asking for a perfect 1code, there is a largish literature. There is no characterization of the regular graphs which contain a perfect 1code, but a useful necessary condition is the that the graph has $1$ as an eigenvalue. The binary Hamming codes provide examples in the $d$cube when $d+1$ is a power of two. 


I think this is possible. First note that it if the graph is disconnected it is trivial. Consider two copies of this graph: Vertices $4$ and $5$ are degree $3$ and all other are $4$. Vertex $3$ is not adjacent to $4$ or $5$. Connect $4$ to $4'$ and $5$ to $5'$ in the other copy to get $4$ regular graph with $3,3'$ having all their neighbourhood in the two copies. The edges:



Is it the case that for each integer $r \geq 2$ the graph $K_{r, r}$ does not admit such a partition? Let $\{V_1, V_2\}$ be a bipartition of $K_{r, r}$. Without loss of generality let $v \in V_1$. Then $v$ and all vertices in $V_2$ must occur in one part of the partition, leaving us with an independent set $V_1  \{v\}$. On the other hand, every complete graph satisfies your property, the partition being the graph itself. 

