jmc

Solving for $a,$ we find $$a=\frac{-x^4+x^2-1}{x^3+x}=-\frac{x^4-x^2+1}{x^3+x}=-\frac{x^2-1+\frac{1}{x^2}}{x+\frac{1}{x}}.$$Let $u=x+\frac{1}{x}.$ Then $u^2=x^2+2+\frac{1}{x^2},$ so $$a=-\frac{u^2-3}{u}.$$If $x$ is positive, then by AM-GM, $u=x+\frac{1}{x}\ge 2.$ Also, $$a+\frac{1}{2}=-\frac{2u^2-u-6}{u}=-\frac{(u-2)(2u+3)}{u}\le 0,$$so $a\le -\frac{1}{2}.$

Furthermore, if $2\le u\le v,$ then $$\begin{array}{rl}-\frac{v^2-3}{v}+\frac{u^2-3}{u}& =\frac{-u{v}^2+3u+u^{2}v-3v}{uv}\\ & =\frac{(u-v)(uv+3)}{uv}\le 0,\end{array}$$which shows that $a=-\frac{u^2-3}{u}=-u+\frac{3}{u}$ is decreasing on $[2,\infty ).$ As $u$ goes to $\infty ,$ $-u+\frac{3}{u}$ goes to $-\infty .$ (Note that $u=x+\frac{1}{x}$ can take on any value that is greater than or equal to 2.)

Similarly, we can show that if $x$ is negative, then $$a=\frac{-x^2+x^2-1}{x^3+x}\ge \frac{1}{2},$$and that $a$ can take on all values greater than or equal to $\frac{1}{2}.$

Hence, the possible values of $a$ are $$a\in \boxed{\left(-\infty ,-\frac{1}{2}\right]\cup \left[\frac{1}{2},\infty \right)}.$$