= Solution
With $q_2=-q_1$, both field contributions between the charges point the same way. Their signed force on $q_3$ is
$$
F_x(f)=\frac{k_eq_1q_3}{L^2}\left[\frac1{(1+f)^2}+\frac1{(1-f)^2}\right]
=\frac{2k_eq_1q_3}{L^2}\frac{1+f^2}{(1-f^2)^2},\qquad k_e=\frac1{4\pi\epsilon_0}.
$$
Since $f$ is dimensionless, the small-displacement expansion is $|f|\ll1$, giving
$$
\boxed{F_x(f)=\frac{2k_eq_1q_3}{L^2}\left(1+3f^2+O(f^4)\right).}
$$
The force at the midpoint is nonzero and the first variation is quadratic, not linear. In particular, the midpoint is not an equilibrium supporting a linear restoring-force approximation. The printed comparison of $f$ with $L$ is dimensionally inconsistent; the expansion above keeps their distinct roles.
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