. . . . . . . . . .
(I have re-labelled the coordinates of the inner triangle, which becomes PQR instead of DEF. Also, L, M and N are the points of trisection on BC, CA and AB.)
The answer is that the area of PQR is one-seventh of the area of ABC.
The only way I can see to prove this is by using vectors. Let a, b and c be position vectors of the points A, B and C. The point L has position vector (1/3)(2b+c), and similarly (1/3)(2c+a) for M, and (1/3)(2a+b) for N.
The point P lies on both BM and CN, so its position vector p must be of the form and also of the form . Solve these equations to see that . Similarly, the position vectors of Q and R are and .
Now think of all these vectors as being in 3-dimensional space (with z-coordinate 0). The standard formula for the area of a triangle is given by a cross product in the form area(ABC) = . If you apply the same formula to the triangle PQR then you get its area as (using the above formulas for p, q and r and simplifying the result. This is 1/7th of the area of ABC.