Geometry with triangle. Any other solutions(advice) are welcome.
$\angle BAC=60^\circ, \;\;\;\angle ACB=x,\;\;\; \overline {BD}=\overline{BC}=\overline{CE}  $

 A: 
$$\angle CBA=120^\circ-x$$
Triangle $DBC$ is isosceles: 
$$\angle CDB=\angle DCB=\frac12\angle CBA=60^\circ-\frac x2$$
Triangle $BCE$ is isosceles: 
$$\angle CEB=\angle CBE=\frac12\angle BCA=\frac x2$$
Now, from triangle:
$$\angle BFC=120^\circ$$
Take a look at quadrilateral $ABFC$: The sum of opposite angles $\angle A$ and $\angle F$ is $180^\circ$ so quadrialteral $ABCD$ is cyclic. Because of that:
$$\angle FAB=\angle FCB=60^\circ-\frac x2=\angle FDB$$ 
So triangle $FAD$ is isosceles and:
$$FA=FD\tag{1}$$
In a similar way:
$$\angle FAC=\angle FBC=\frac x2=\angle FEA$$ 
So triangle $FEA$ is isosceles and:
$$FA=FE\tag{2}$$
From (1) and (2) you have:
$$FD=FE$$ 
...and triangle $FDE$ must be isosceles. Angle $\angle DFE=120^\circ$ and therefore:
$$\angle FDE=\angle FED=30^\circ$$
It follows that:
$$\angle DEA=\angle DEF+\angle FEA=30^\circ+\frac x2$$
A: Drawing the rhombus is a nice way to start but we can finish differently.
As soon as you find that $\triangle BDF$ is equilateral, you know that
$FD = FB = FE.$ That is, the points $D,$ $B,$ and $E$ all lie on the same circle with center at $F.$
Referring to this same circle, since the central angle $\angle BFD$ is $60^\circ,$
the inscribed angle $\angle BED$ is $30^\circ.$
We have $\angle CEF = \angle ACB = x$ and $EB$ is the angle bisector of
$\angle CEF,$ so $\angle CEB = \frac12 \angle CEF = \frac 12x.$
Then $\angle CED = \angle CEB + \angle BED = \frac 12x + 30^\circ.$
