How to show that $\int_{0}^{\infty}{(2x)^4\over (1-x^2+x^4)^4}\mathrm dx=\int_{0}^{\infty}{(2x)^4\over (1-x^2+x^4)^3}\mathrm dx=3\pi$

Consider the transformation $x=1/y$. Then

$$ J=16\int_0^{\infty} dy\frac{1}{y^2} \frac{1/y^4}{(1-1/y^2+1/y^4)^3}=8\int_{-\infty}^{\infty} dy\frac{1}{(y^2-1+1/y^2)^3}=\\8\int_{-\infty}^{\infty} dy\frac{1}{((y-1/y)^2+1)^3}\underbrace{=}_{(\star)}8\int_{-\infty}^{\infty}dz\frac{1}{(z^2+1)^3}=3\pi $$

and your proof is complete (for the next to last equality $(\star)$ we applied Glasser's Master theorem which is not difficult to proof for this special case)

Furthermore applying the same substitution $x=1/y$ again, it is a matter of straightforward algebra that

$$ \Delta=J-I=-\Delta $$

so

$$ \Delta=0 \,\,\text{or}\,\,I=J=3\pi $$

QED

$\begin{align}I&=\int_{0}^{\infty}{(2x)^4\over (1-x^2+x^4)^4}\mathrm dx \qquad\\ &=\int_{0}^{\infty} \dfrac{16x^4(1+x^2)^4}{(1+x^6)^4} dx\\ &=\int_{0}^{\infty} \dfrac{16\cdot {{x}^{12}}+64\cdot {{x}^{10}}+96\cdot {{x}^{8}}+64\cdot {{x}^{6}}+16\cdot {{x}^{4}}}{(1+x^6)^4} dx\\ \end{align}$

Perform the change of variable $y=x^6$,

$\begin{align} I&=\int_{0}^{\infty} \dfrac{\tfrac{8}{3}x^{\tfrac{7}{6}}+\tfrac{32}{3}x^{\tfrac{5}{6}}+16x^{\tfrac{1}{2}}+\tfrac{32}{3}x^{\tfrac{1}{6}}+\tfrac{8}{3}x^{-\tfrac{1}{6}}}{(1+x)^4} dx\\ &=\dfrac{8}{3}\text{B}\left(\dfrac{13}{6},\dfrac{11}{6}\right)+\dfrac{32}{3}\text{B}\left(\dfrac{13}{6},\dfrac{11}{6}\right)+16\text{B}\left(\dfrac{3}{2},\dfrac{5}{2}\right)+\dfrac{32}{3}\text{B}\left(\dfrac{7}{6},\dfrac{17}{6}\right)+\dfrac{8}{3}\text{B}\left(\dfrac{5}{6},\dfrac{19}{6}\right)\\ &=\dfrac{8}{3}\times \dfrac{35\pi}{648}+\dfrac{32}{3}\times \dfrac{35\pi}{648}+16\times \dfrac{\pi}{16}+ \dfrac{32}{3}\times\dfrac{55\pi}{648}+\dfrac{8}{3}\times \dfrac{91\pi}{648}\\ &=\boxed{3\pi} \end{align}$

Addendum:

$B$ is the Euler beta function.

$\begin{align} \text{B}\left(\dfrac{13}{6},\dfrac{11}{6}\right)&=\dfrac{\Gamma\left(\dfrac{13}{6}\right)\Gamma\left(\dfrac{11}{6}\right)}{\Gamma\left(4\right)}\\ &=\dfrac{1}{6} \left(\dfrac{7}{6}\Gamma\left(\dfrac{7}{6}\right)\right)\times \left(\dfrac{5}{6}\Gamma\left(\dfrac{5}{6}\right)\right)\\ &=\dfrac{1}{6} \left(\dfrac{7}{6}\times\dfrac{1}{6}\Gamma\left(\dfrac{1}{6}\right)\right)\times \left(\dfrac{5}{6}\Gamma\left(\dfrac{5}{6}\right)\right)\\ &=\dfrac{35}{1296}\Gamma\left(\dfrac{1}{6}\right)\Gamma\left(\dfrac{5}{6}\right)\\ &=\dfrac{35}{1296}\times \dfrac{\pi}{\sin\left(\pi\times \tfrac{1}{6}\right)}\\ &=\boxed{\dfrac{35\pi}{648}} \end{align}$