Reminder of how we defined the relation:
\(f \thicksim g\) if and only if there is a \(k \in \mathbb{R}\) such that f(x) = g(x) for every \(x \geq k\).
First, let's check that your solution has the right outline and has the easy parts working.
Reflexive
Let f be a function from the reals to the reals. Pick k=0. (Any value will work.) Then f(x) = f(x) for every \(x \geq k\). So \(f \thicksim f\).
Symmetric
Let f and g be functions from the reals to the reals. Suppose that \(f \thicksim g\). Then, by the definition of \(\thicksim\), there is a real number k such that f(x) = g(x) for every \(x \geq k\). But then g(x) = f(x) for every \(x \geq k\). So we must have \(g \thicksim f\).
Transitive
Let f, g, and h be functions from the reals to the reals. Suppose that \(f \thicksim g\) and \(g \thicksim h\).
... here a miracle happens ...
So we must have \(f \thicksim h\). Since our relation is reflexive, symmetric, and transitive, it is an equivalence relation.
If your solution looks like the above, move on to the full solutions. But, first, some hints in case you're stuck on filling in the "miracle" part.
Remember to apply the definition of the relation, to reduce the gap between the starting information and your goal. Notice that the relation introduces a local variable k. So it's like the definition of "divides": you have to use a fresh variable name when you expand the definition a second (or third) time.
It may help to draw a picture of three functions, where the bound k is different for the two pairs. That is, f and g start to differ at some point, but h branches off from g at a different point. What bound can you use when comparing f to h?