$1.37 \sqrt[3]{\frac{2 x^{2}}{9+18 x+9 x^{2}}} \cdot \sqrt{\frac{(1+x) \sqrt[3]{1-x}}{x}} \cdot \sqrt[3]{\frac{3 \sqrt{1-x^{2}}}{2 x \sqrt{x}}}$.
1.37 Let's write all three factors under one common root:

$$
\begin{aligned}
& \sqrt[6]{\frac{4 x^{4}(1+x)^{3} \cdot 9(1-x^{2})(1-x)}{9^{2}(1+2 x+x^{2})^{2} x^{3} \cdot 4 x^{3}}}=\sqrt[6]{\frac{(1+x)^{4}(1-x)^{2}}{9 x^{2}(1+x)^{4}}}= \\
& =\sqrt[6]{\frac{(1-x)^{2}}{9 x^{2}}}=\sqrt[3]{\left|\frac{1-x}{3 x}\right|}
\end{aligned}
$$

Since $x \in (0,1]$, we get the answer: $\sqrt[3]{\frac{1-x}{3 x}}$.

Answer: $\sqrt[3]{\frac{1-x}{3 x}}$