Punoquillads is on the right track, though I think he needs to relabel the points opposite the vertices, since DEF is taken by the triangle in the center. Call the point G where the line from vertex A meets the opposite side BC; similarly, point H is on the side opposite vertex B, and I is opposite C.
For an explanation that avoids barycentric coordinates, consider all the “one third” sized triangles (ABG, BCH, and CAI). Each of these have area 1/3 that of ABC, so taken separately and added they will have an area equal to ABC. Of course they don’t match ABC because they overlap; the three smaller triangles attached to each vertex (ADI, BEG, and CFH) are all counted twice in the summation. Consequently, the sum of area of these three small triangle must equal the area of the interior triangle DEF.
If you assume ABC is equilateral (against the rules, I know), it should be obvious in this case that all the little corner triangles are congruent. By carefully looking at the angles formed by the lines, it’s fairly easy to prove each little corner triangle is similar (though flipped) to the larger “one third size” triangle it shares its most acute vertex with (i.e. ADI ~ ABG, BEG ~ BCH, CFH ~ CHI). Also, these littler triangles have sides which are 1/3 that of their larger similars (i.e. AI = 1/3(AB), BG = 1/3(BC), CH=1/3(CA)), so they must be 1/9 the area of the larger similar, which themselves are 1/3 the size of ABC. Consequentially, the littler triangles have area 1/27(ABC). Adding the three of them up and noting that together their area equals the area of the central triangle DEF, this means the area of DEF must be 1/9(ABC).
Now you could argue that your teachers wouldn’t assign the problem if it didn’t have a single solution for all such triangles, so this is a rather sneaky “proof” that DEF = 1/9(ABC). I’d love to find something more general though…