Show that if $G$ is a $k$-regular graph of order $n$, then G has $\frac{nk}{2}$ edges.
My try:
By hipostesis we know that for all $n$ vertices there are $k$ edges impacting on each of them. Then the total of edges impacting on all vertices is equal to $kn$ and by definition, the total of edges of $G$ is half of all the impacting edges. Then the total of edges is $\frac{nk}{2}$
I don´t know if it´s correct.
It looks pretty much good. A couple notes:
In English, the word generally used is "hypothesis", not "hipostesis".
The phrase "impacting on" is sort of difficult to understand. I would generally say "incident to" -- an edge is said to be incident to a vertex if that vertex is one of the endpoints.
It would be clearer instead of using the phrase "total number", which can be somewhat ambiguous, to say something like: "The total number of pairs $(v,e)$ where $e$ is an edge incident to vertex $v$ is $kn$."
"By definition" isn't a terribly good reason why you're dividing by $2$ -- it would be clearer if you said something like "as each edge is incident to exactly two vertices..."
Overall, however, your math is perfectly correct.