I'm trying to prove a construction by induction that for a planar graph, the maximum number of edges is $3v-6<=e$.
The construction suggested to me from this diagram:
from here: Formal construction of the maximum planar graph of order n.
basis $v=3$: $3*3-6<=3$. check.
step: here I have some troubles, I get some things mixed is that what I should prove: $3(k+1)-6<=m$?
As I understand it, you attempting to prove that there is a planar graph for any $v\ge 3$ with the maximal number of edges, $e=3v-6$.
Your base case is correct: we can just exhibit the triangle graph and check that it meets the desired formula.
For the inductive step, we need to go from our inductive hypothesis that there is a graph $G$ for $v\ge 3$ such that $e=3v-6$, and prove that we can construct a graph $G'$ on $v{+}1$ vertices with $e=3(v+1)-6 = 3v-3$.
The figure suggests the method here. If we examine the planar graph $G(V,E)$ we see that the face (region) outside the graph is bounded by $3$ edges. To make $C'$, we can add our extra vertex outside the existing boundary of $G$ and connect to all three boundary points, meaning that we now have $|E|+3=3v-6+3 = 3v-3$ edges as required.