Use the printed half-unit initial value in a derivative recurrence

$f^{\prime\prime}=f^{\prime}\ln(\sec(ax))+af\tan(ax)$. Multiply by $f$ and use $f\ln(\sec(ax))=f^{\prime}$ to obtain $ff^{\prime\prime}=(f^{\prime})^2+af^2\tan(ax)$, as required.

Differentiation gives

$\begin{aligned}&f f^{\prime\prime\prime}-f^{\prime}f^{\prime\prime}\\ & =a^2f^2\sec^2(ax)\\ &{}+2aff^{\prime}\tan(ax).\end{aligned}$

At zero, $f=1/2$, $f^{\prime}=0$, $f^{\prime\prime}=0$, so $\tfrac12 f^{\prime\prime\prime}(0)=a^2/4$, giving $f^{\prime\prime\prime}(0)=a^2/2$. Thus $f(x)=1/2+a^2x^3/12+O(x^5)$. The rendered source explicitly gives $f(0)=1/2$; native extraction loses its denominator. The printed series is consistent with the visual source, not an erroneous half of a unit-initial-value solution.