First note that $x = \sqrt[3]{2\sqrt[3]{2x-1}-1}$. Let $f(x) = \sqrt[3]{2x-1}$. We then have $x = f(f(x))$. First, we observe that if $x_0 = f(x_0)$, then $x_0 = f(f(x_0)$. Note that $f$ is an increasing function. Given this, if $x_0 > f(x_0)$, then $f(x_0) > f(f(x_0))$ and conversely $x_0$ < $f(x_0) \Rightarrow f(x_0)$ < $f(f(x_0))$. So we must have $x_0 = f(x_0)$. Given this we must just find $x = \sqrt[3]{2x-1}$ or $x^3 - 2x - 1 = 0$. Since $x=1$ is clearly a root, we factor as $(x-1)(x^2 + x + 1)$. The quadratic second term has solutions $(-2+-\sqrt{5})/2$.