Other ways to evaluate $\lim_{x \to 0} \frac 1x \left [ \sqrt[3]{\frac{1 - \sqrt{1 - x}}{\sqrt{1 + x} - 1}} - 1\right ]$? Using the facts that:
$$\begin{align}
\sqrt{1 + x} &= 1 + x/2 - x^2/8 + \mathcal{o}(x^2)\\
\sqrt{1 - x} &= 1 - x/2 - x^2/8 + \mathcal{o}(x^2)\\
\sqrt[3]{1 + x} &= 1 + x/3 + \mathcal{o}(x)
\end{align}$$
I was able to evaluate the limit as follows:
$$\begin{align}
\lim_{x \to 0} \frac 1x \left [ \sqrt[3]{\frac{1 - \sqrt{1 - x}}{\sqrt{1 + x} - 1}} - 1\right ] &\sim \lim_{x \to 0} \frac 1x \left [ \sqrt[3]{\frac{\dfrac x2 + \dfrac{x^2} 8}{\dfrac x2 + \dfrac{x^2} 8}} - 1\right ] =\\
&= \lim_{x \to 0} \frac 1x \left [ \sqrt[3]{1 + \frac{2x^2}{4x - x^2}} - 1\right ] \sim\\
&\sim \lim_{x \to 0} \frac{2x^2}{12x^2 - 3x^3} = \frac 16
\end{align}$$
What are other ways to evaluate it? Maybe pure algebraically? I tried to rationalize the denominator, but got stuck at some point...

Using:
$$
   1 -\sqrt{1-x} = \frac{x}{1+\sqrt{1-x}}\qquad \frac{1}{\sqrt{1+x}-1} = \frac{\sqrt{1+x}+1}{x}
$$
we have:
$$
  \frac{1 -\sqrt{1-x}}{\sqrt{1+x}-1} = \frac{\sqrt{1+x}+1}{1+\sqrt{1-x}} = 1 + \frac{\sqrt{1+x}-\sqrt{1-x}}{1+\sqrt{1-x}} = 1 + \underbrace{\frac{2x}{\left(\sqrt{1+x}+\sqrt{1-x}\right)\left(1+\sqrt{1-x}\right)}}_{w}
$$
And using
$$
    \sqrt[3]{1+w} - 1 = \frac{w}{1 + \sqrt[3]{1+w} + \sqrt[3]{\left(1+w\right)^2} }
$$
we have
$$
  \lim_{x \to 0}\frac{1}{x} \left(\sqrt[3]{ \frac{1 -\sqrt{1-x}}{\sqrt{1+x}-1} } - 1\right) = \lim_{x \to 0} \frac{w}{x} \cdot \frac{1}{1 + \sqrt[3]{1+w} + \sqrt[3]{\left(1+w\right)^2} } = \frac{2}{4} \cdot \frac{1}{3} = \frac{1}{6}
$$
