Solution:
Answer: $\frac{\sqrt[3]{28}}{3}$. Cubing the given equation yields
$$1=\left(1-x^{3}\right)+3 \sqrt[3]{\left(1-x^{3}\right)\left(1+x^{3}\right)}\left(\sqrt[3]{1-x^{3}}+\sqrt[3]{1+x^{3}}\right)+\left(1+x^{3}\right)=2+3 \sqrt[3]{1-x^{6}}$$
Then $\frac{-1}{3}=\sqrt[3]{1-x^{6}}$, so $\frac{-1}{27}=1-x^{6}$ and $x^{6}=\frac{28}{27}$ and $x^{2}=\frac{\sqrt[3]{28}}{3}$.