ABC and CDE are equilateral triangles, as the figure shows.

Since nothing is given about where E point lies on the line AC, we can assume E to be the mid point of AC to solve the question for a specific case. Why this should work is because no other measurements are given, and this means, the angle x should be same for every position of E on line AC.

So solving for this specific case:
Line BE will bisect the angle B and will be perpendicular to AC.
since we have assumed E to be the mid point of AC, EC=AE. Moreover, since triangle CDE is equilateral, EC = DE.
So triangle ADE becomes isosceles as AE = DE ( = EC)

Since \(\angle AED\) = 120° (as \(\angle AED\) + \(\angle DEC\) = 180°, and \(\angle DEC\) = 60°),
the \(\angle DAE\) and \(\angle EDA\) will both be 30°
Now, as I mentioned earlier, BE is perpendicular to AC, so \(\angle AEF\) = 90°
In triangle AEF, we know two angles, we can find the third.
\(\angle FAE\) + \(\angle AEF\) + \(\angle x\) = 180°
30° + 90° + \(\angle x\) = 180°
\(\angle x\) = 60°
So answer is D