Three, (20 points) Solve the equation
$$
\left(\frac{x}{x-1}\right)^{2}+\left(\frac{x}{x+1}\right)^{2}=\frac{40}{9} .
$$
Three, removing the denominators, we get
$$
x^{2}(x+1)^{2}+x^{2}(x-1)^{2}=\frac{40}{9}\left(x^{2}-1\right)^{2},
$$

After simplification, we have $x^{2}\left(x^{2}+1\right)=\frac{20}{9}\left(x^{2}-1\right)^{2}$. Let $y=x^{2}$, the original equation becomes $y(y+1)=\frac{20}{9}(y-1)^{2}$, which simplifies to $11 y^{2}+49 y+20=0 \Rightarrow(y-4)(11 y-5)=0$. Solving this, we get $y=4$ or $y=\frac{5}{11}$. Therefore, $x= \pm 2$ or $x= \pm \frac{\sqrt{55}}{11}$.
Upon verification, $x= \pm 2, x= \pm \frac{\sqrt{55}}{11}$ are the roots of the original equation.