jmc

If $x=0,$ then the expression is equal to 0, so assume that $x\ne 0.$ Then dividing the numerator and denominator by $x^4,$ we get $$\frac{1}{x^4+2x^2-4+\frac{8}{x^2}+\frac{16}{x^4}}.$$By AM-GM, $$x^4+\frac{16}{x^4}\ge 2\sqrt{x^4\cdot \frac{16}{x^4}}=8,$$and $$2x^2+\frac{8}{x^2}\ge 2\sqrt{2x^2\cdot \frac{8}{x^2}}=8,$$so $$\frac{1}{x^4+2x^2-4+\frac{8}{x^2}+\frac{16}{x^4}}\le \frac{1}{8+8-4}=\frac{1}{12}.$$Equality occurs when $x=\sqrt{2},$ so the maximum value is $\boxed{\frac{1}{12}}.$