An equation has the following form:
$$Ae^{-i D x} + Be^{-i E x} = Ce^{-i F x}$$
where $A,B,C \in \mathbb{C}$ and $D,E,F \in \mathbb{R}$ are all constants and $x \in \mathbb{R}$, while $i$ is the imaginary unit. These constants should be determined in order to make the equation true for any $x$. Which is the correct approach?
Taking the derivative with respect to $x$ of the whole function and making it $0$ is somewhat confusing:
$$\frac{d}{dx} [Ae^{-i D x} + Be^{-i E x} - Ce^{-i F x}] = 0$$
$$-iD Ae^{-i D x} - iE Be^{-i E x} + iF Ce^{-i F x} = 0$$ $$D Ae^{-i D x} + E Be^{-i E x} - F Ce^{-i F x} = 0$$
Maybe the simplest solution (watching directly the equation) can be
$$D = E = F$$
so that all the exponential (which are never 0) simplify. But is this the only way to make true the initial equation for any real $x$?
Sorry for the late answer, I could find time just recently. It turns out your approach is more useful than what I suggested, and by considering the first and second derivatives one reaches a definitive answer. One argument I have used extensively is this: If we want the equation to be satisfied by any real $x$, then it must be satisfied by a particular $x$ too. If one chooses $x$ wisely this becomes very practical. So even though my answer is long, the solution is simply by considering derivatives and plugging in particulars. Also in case you are interested in (at least) algebra your question is related to exponential polynomials, and Google provides some papers (e.g. the first hit is Ritt's "On the Zeros of Exponential Polynomials").
We would like to find all possible combinations of $A,B,C\in\mathbb{C}$ and $D,E,F\in\mathbb{R}$ for which any real $x$ satisfies the equation
$$\label{0}Ae^{-iDx}+Be^{-i Ex}-Ce^{-i Fx}=0\tag{0}.$$
Considering the first and second derivatives of \ref{0}, we have
$$\label{1}-iADe^{-iDx}-iBEe^{-i Ex}+iCFe^{-i Fx}=0,\tag{1}$$
$$\label{2}-AD^2e^{-iDx}-BE^2e^{-i Ex}+CF^2e^{-i Fx}=0.\tag{2}$$
In a more convenient form, we have the following system of equations (observe that we can actually get as many equations as we want where the powers of $D,E$ and $F$ are always equal by differentiating):
\begin{align} &Ae^{-iDx}&+&Be^{-i Ex}&-&Ce^{-i Fx}&=0\tag{0}\\ &ADe^{-iDx}&+&BEe^{-i Ex}&-&CFe^{-i Fx}&=0\tag{1}\\ &AD^2e^{-iDx}&+&BE^2e^{-i Ex}&-&CF^2e^{-i Fx}&=0\tag{2}\\ \end{align}
Suppose $A,B,C,D,E,F$ are constants for which \ref{0} holds for arbitrary $x$. Then the system \ref{0}-\ref{2} must hold for $x$ also, and in particular the system must hold for $x:=0$. Plugging this value in the system, we obtain
\begin{align} &\label{0*}A&+&B&-&C&=0\tag{0*}\\ &\label{1*}AD&+&BE&-&CF&=0\tag{1*}\\ &\label{2*}AD^2&+&BE^2&-&CF^2&=0\tag{2*} \end{align}
First equation gives that $C=A+B$. Substituting $C$ accordingly in the second equation we obtain $A(D-F)+B(E-F)=0$. Then we have five cases:
In the first case $C=0$ and $D,E,F$ are arbitrary. In the second case $C=B$ and $D$ is arbitrary, and similarly in the third case $C=A$ and $E$ is arbitrary. For the fourth case we still have $C=A+B$ while any real $D$ works.
Finally we have the fifth case left. Doing the substitution $C=A+B$ in \ref{2*} and rearranging we obtain
$$A(D-F)(D+F)+B(E-F)(E+F)=0.$$
Since $A(D-F)=-B(E-F)$, have $A(D-F)[(D+F)-(E+F)]=0$, i.e.,
$$A(D-F)(D-E)=0.$$
The first two factors on LHS is nonzero, so $D=E$. Rearranging \ref{0} accordingly, we obtain
$$C(e^{-iDx}-e^{-iFx})=0.$$
Then either $C=0$, in which case $A=-B$ are arbitrary complex numbers and $D=E,F$ are arbitrary real numbers, or else
$$\label{*}e^{-iDx}=e^{-iFx}.\tag{*}$$
If $D$ is zero, $F\neq0$ by the hypotheses of the fifth case. But then solving the equation results in countably many solutions $x$. To be precise $x$ is forced to be of the form $\dfrac{2n\pi}{F}$ for some integer $n$, $\Large\unicode{x21af}$. Similarly $F$ is nonzero. Then plugging $x:=\dfrac{\pi}{D}$ and $x:=\dfrac{\pi}{F}$ in \ref{*}, we have that $F=(2k+1)D$ and $D=(2l+1)F$ for some nonzero integers $k,l$, respectively. Then
$$D=(2k+1)(2l+1)D\implies 1=(2k+1)(2l+1).$$
Since the factors on RHS are integers that are nonunital, both of them must be $-1$. Consequently $F=-D$. But this forces $x$ to be of the form $\dfrac{n\pi}{D}$ for some integer $n$, $\Large\unicode{x21af}$. Hence \ref{*} is impossible.
(TL;DR) In conclusion if any real $x$ satisfies \ref{0}, then $A,B,C,D,E,F$ must satisfy one of the following restrictions:
Conversely it is clear that any $6$-tuple $A,B,C,D,E,F$ satisfying one of these five conditions will render \ref{0} having any $x\in\mathbb{R}$ as a solution.