There are real numbers $a$ and $b$ such that for every positive number $x$, we have the identity
\[
  \tan^{-1} \left( \frac{1}{x} - \frac{x}{8} \right) + \tan^{-1}(ax) + \tan^{-1}(bx) 
    = \frac{\pi}{2} \, .
\](Throughout this equation, $\tan^{-1}$ means the inverse tangent function, sometimes written $\arctan$.)  What is the value of $a^2 + b^2$?
By the tangent addition formula,
\[\tan (\tan^{-1} p + \tan^{-1} q) = \frac{p + q}{1 - pq}.\]From the given equation,
\[\tan^{-1} ax + \tan^{-1} bx = \frac{\pi}{2} - \tan^{-1} \left( \frac{1}{x} - \frac{x}{8} \right).\]Then
\[\tan (\tan^{-1} ax + \tan^{-1} bx) = \tan \left( \frac{\pi}{2} - \tan^{-1} \left( \frac{1}{x} - \frac{x}{8} \right) \right),\]The left-hand side is $\frac{ax + bx}{1 - abx^2}.$  The right-hand side is
\begin{align*}
\tan \left( \frac{\pi}{2} - \tan^{-1} \left( \frac{1}{x} - \frac{x}{8} \right) \right) &= \frac{1}{\tan \left( \tan^{-1} \left( \frac{1}{x} - \frac{x}{8} \right) \right)} \\
&= \frac{1}{\frac{1}{x} - \frac{x}{8}} \\
&= \frac{x}{1 - \frac{1}{8} x^2}.
\end{align*}Hence, $a + b = 1$ and $ab = \frac{1}{8},$ so
\[a^2 + b^2 = (a + b)^2 - 2ab = 1 - \frac{2}{8} = \boxed{\frac{3}{4}}.\]