it's Clobberin time

$f(x) = \dfrac {x^4}{x^8+2x^6-4x^4+8x^2+16}$

We note that both $x^4 \ge 0$ and $x^8+2x^6-4x^4+8x^2+16 > 0$ are even, therefore, $f(x)$ is even and positive and the least value is $f(0) = 0$ when $x^4 = 0$ .

$\begin{aligned} f(x) & = \dfrac {x^4}{x^8+2x^6-4x^4+8x^2+16} \\ & = \dfrac{1}{x^4+2x^2-4+\frac{8}{x^2}+\frac{16}{x^4}} \\ & = \dfrac{1}{\left( x^2 + \frac{4}{x^2}\right)^2 - 8 + 2\left( x^2 + \frac{4}{x^2}\right) - 4} \\ & = \dfrac{1}{\left( x^2 + \frac{4}{x^2}\right)^2 + 2\left( x^2 + \frac{4}{x^2}\right) - 12} \\ & = \dfrac{1}{\left( x^2 + \frac{4}{x^2}\right)^2 + 2\left( x^2 + \frac{4}{x^2}\right) +1 - 13} \\ & = \dfrac{1}{\left( x^2 + 1 + \frac{4}{x^2}\right)^2 - 13} \end{aligned}$

$f(x)$ is maximum, when $x^2 + 1 + \frac{4}{x^2}$ is minimum.

$\dfrac {d}{dx} \left( x^2 + 1 + \dfrac{4}{x^2}\right) = 2x - \dfrac{8}{x^3}$

Equating to $0\quad \Rightarrow 2x - \dfrac{8}{x^3} = 0 \quad \Rightarrow 2x^4 = 8 \quad \Rightarrow x^4 = 4 \quad \Rightarrow x = \sqrt{2}$

The greatest value of $f(x)$ is:

$\begin{aligned} f(\sqrt{2}) & = \dfrac{1}{\left( (\sqrt{2})^2 + 1 + \frac{4}{(\sqrt{2})^2}\right)^2 - 13} \\ & = \dfrac {1}{(2+1+2)^2 -13} \\ & = \dfrac{1}{12} = \boxed{0.08333} \end{aligned}$