route finding is a trivial task that commonly appears in any beginner-level course of programming and even matrixes applications of 1st degree math courses
Nope - some simplifed algorithms are taught. And even in this case, the examples (and practically computable applications) are, for a good reason, limited to small two-dimanesional problems. You can even see this working in ED: whenever I plot a longer route, every ten full range jumps or so I get a small jump in the plotted route, even though there would be further suitable stars en route. Indicating that I'm crossing one of the segment boundaries the plotter has introduced to split the route into manageable segments.
you dont have to check billions of routes

sorry - but that one actually made me laugh (and reach for a pencil). Sol - Colonia is about 20,000 ly. To reach the Neutron layer, you'll need to deviate by about 1,000 ly from the direct route. So, that's roughly speaking a volume of 2,000 x 2,000 x 20,000 ly³ - 8e10 ly³. Yes, you could reduce it a little bit - but that's an order-of-magnitude estimation, let's not haggle about a factor 10. To make calculation simpler, I assume on average one star per 20 cubic lightyears (something like 8 ly to the nearest neighbour, more on the rim, much less in the core). Divide 8e10 by 20, and you end up with 4e9 stars in that volume. Roughly 1% of the number of stars in the Galaxy, looks about right.
So, how many routes can you plot through 4 billion stars in three dimensions? The point of the algorithms you mentioned above is to reduce that selection - but even using those, "billions" is probably lowballing the number of computationally feasible routes by "a few" orders of magnitude. And remember - you wanted the fastest (in time) overall route.
there is a lot of optimizations techniques and tricks about route finding and in our case it's absolutelly solved problem
Errmmm... no. There are a few widely used algorithms that provide solutions. For a small number of nodes, a shortest path is computationally calculable - but for a large number of nodes (in two dimensions!), V*
log(V) is the best these algorithms will do (still better than V²), and that always catches you out. For large numbers of nodes, there are approximations. FD uses one of them, Spansh is likely using another.
it's already done by Spansh plotter
Again, no. Spansh may use a different algorithm, but, as already stated, more important is that Spansh only uses a tiny subset of the stars in the Galaxy. And people who really want to push the edge (like breaking the Sol-Colonia record, or going to Colonia without a fuel scoop) still use their own algorithms.