6. Let the function be
$$
f(x)=\frac{x(1-x)}{x^{3}-x+1}(0<x<1) \text {. }
$$

If the maximum value of $f(x)$ is $f\left(x_{0}\right)$, then $x_{0}=$ $\qquad$ .
$$
\begin{array}{l}
\text { 6. } \frac{\sqrt{2}+1-\sqrt{2 \sqrt{2}-1}}{2} . \\
\text { Let } f^{\prime}(x)=\frac{x^{4}-2 x^{3}+x^{2}-2 x+1}{\left(x^{3}-x+1\right)^{2}} \\
=\frac{\left(x^{2}-(\sqrt{2}+1) x+1\right)\left(x^{2}+(\sqrt{2}-1) x+1\right)}{\left(x^{3}-x+1\right)^{2}}=0 .
\end{array}
$$

Then $\left(x^{2}-(\sqrt{2}+1) x+1\right)\left(x^{2}+(\sqrt{2}-1) x+1\right)=0$.
Since the discriminant of $x^{2}+(\sqrt{2}-1) x+1=0$ is less than 0, this equation has no real roots.
For $x^{2}-(\sqrt{2}+1) x+1=0$, by
$$
\frac{\sqrt{2}+1+\sqrt{2 \sqrt{2}-1}}{2}>1 \text {, }
$$

we know that this equation has a unique real root
$$
x_{0}=\frac{\sqrt{2}+1-\sqrt{2 \sqrt{2}-1}}{2} \in(0,1) \text {. }
$$

When $0<x<x_{0}$, $f^{\prime}(x)>0$, $f(x)$ is strictly increasing;

When $x_{0}<x<1$, $f^{\prime}(x)<0$, $f(x)$ is strictly decreasing.
Therefore, when $x=x_{0}$, $f(x)$ achieves its maximum value.
Thus, $x_{0}=\frac{\sqrt{2}+1-\sqrt{2 \sqrt{2}-1}}{2}$.