Numerical verification of solution. I have the non-linear equation
\begin{align}
& \left( -\frac{1}{4}\left({\frac { \left( 4\,{x}^{3}+2\,
ex \right) ^{2}}{ \left( {x}^{4}+e{x}^{2}+f \right) ^{3/2}}}\right)+\frac{1}{2}\left({
\frac {12\,{x}^{2}+2\,e}{\sqrt {{x}^{4}+e{x}^{2}+f}}} \right)\right) \\ &+\frac{1}{2} \left({
\frac {x \left( 4\,{x}^{3}+2\,ex \right)  \left( 8\,a{x}^{2}-8\,a+1
 \right) }{\sqrt {{x}^{4}+e{x}^{2}+f}}} \right) \\ &-\quad 4\left( 4\,a{x}^{2}+{a}^{2}
+C-2\,a \right) \sqrt {{x}^{4}+e{x}^{2}+f}=0 
\end{align}
where $x \in  [-7,15]$ and $a \in \mathbb{Q}$. I want to solve this one numerically such that this equation is fulfilled for arbitrary $x \in [-7,15]$ and one parameter $a$ that is entered a priori. So, essentially I want to determine $f,e,C$ such that this equation is fulfilled for the chosen $a \in \mathbb{R}$ and all $ x \in [-7,15]$. Hence, $f,e,C \in \mathbb{Q}$ must be constant and therefore independent of $x$.
 A: First. I want to mention a naive brute-force numerical scheme. If you are looking to solve a set of equations $f_x(a,e,f,c)=0$ for a continuous range of $x$, you can always try picking three values of $x$ at random from $[-1,1]$, and, given $a$, try to solve the three equations $f_x(a,e,f,c)=0$ in three unknowns $e,f,c$ numerically. The numerical solver might not always succeed, but you can try many different starting points for $e,f,c$ at random, and pick different values of $x$. I actually tried this scheme, and it kind of works, finding two or three of the analytical solutions that you gave. I used the default FindRoot solver in Mathematica with no special options at all.
Second. If you multiply your equation by $(x^4+ex^2+f)^{3/2}$ (which is nonzero for almost all $x$, so this is fine), you get the following polynomial in $x$ that has to vanish identically:
$$ -e f+8 a f^2-4 a^2 f^2-4 c f^2+\left(-6 f+2 e f+8 a e f-8 a^2 e f-8 c e f-16 a f^2\right) x^2\\+\left(-3 e+e^2-4 a^2 e^2-4 c e^2+8 f-8 a^2 f-8 c f-24 a e f\right) x^4\\+\left(-2+6 e-8 a e-8 a^2 e-8 c e-8 a e^2-16 a f\right) x^6+\left(4-8 a-4 a^2-4 c-8 a e\right) x^8. $$
All the coefficients of $x$ therefore also have to vanish, and all these coefficients are polynomials $q_k$ in $a, e, f, c$.
Someone better familiar with algebraic geometry can explain this better than I can, but I'll try. The set of all $a, e, f, c$ that make all $q_k$ vanish simultaneously is called the ideal generated by the $q_k$'s. Sage can help here. This ideal will have a Groebner basis, which is a set of "simpler" polynomials that have precisely the same roots. It will also have a "primary decomposition" into "primary ideals", which gives a set of sets of polynomials such that each solution to the original system of polynomial equations solves one of the sets of polynomial equations in the primary decomposition.
The primary ideals are also guaranteed to have some mathematical properties that I find mysterious. It's like if you want to solve $(x-1)(y-1)(z-1)=0$, then the three primary ideals will be $\{x-1\}$, $\{y-1\}$ and $\{z-1\}$.
The point of this is that when I tried to solve your problem in sage, it gave me the following primary decomposition of the ideal:
x,a,e,f,c = QQ['x,a,e,f,c'].gens()
P = (-1 + x**2)*(e*f + 6*f*x**2 + 3*e*x**4 + 2*x**6) + \
  x**2*(e + 2*x**2)*(f + e*x**2 + x**4)*(1 + 8*a*(-1 + x**2)) - \
  4*(f + e*x**2 + x**4)**2*(a**2 + c + a*(-2 + 4*x**2))
PC = [P.coefficient({x:k}) for k in range(0, 1+P.degree(x), 2)]
I = ideal(PC)
I.primary_decomposition()

[Ideal (f, e + 1, a^2 + c - 1)
   of Multivariate Polynomial Ring in x, a, e, f, c over Rational Field,
 Ideal (8*a*f - 16*e*c - 32*f*c - 4*a - e - 9, e^2 - 4*f,
        a*e + 4*e*c + 8*f*c + a + 2, a^2 - 8*e*c - 16*f*c + c - 5,
        64*f^2*c - 32*e*c - 128*f*c - 8*a - 10*e + 20*f - 19,
        16*e*f*c + 8*e*c + 48*f*c + 2*a + 5*e + 5)
   of Multivariate Polynomial Ring in x, a, e, f, c over Rational Field]

Now all that's left is the two solve each set of polynomial equations in $e,f,c$ in terms of $a$, so there is no need for a numerical approach any more.
The two primary ideals it returned correspond exactly to the three analytical solutions you mentioned in your post ($f=0,e=-1,c=1-a^2$ and the two solutions to quadratic equations). If sage's calculation is correct, this is also a proof that there are no other solutions to your problem, so there is no point in looking for them. Indeed, I found none when I used the brute-force numerical scheme.
