Limit in form of 0/0 
If I multiply by  I get zero/2x anyway.
What manipulation needed to get 2/3?
 A: You need to multiply by $$\left(\sqrt[3]{1+x}\right)^2+\left(\sqrt[3]{1+x}\right)\left(\sqrt[3]{1-x}\right)+\left(\sqrt[3]{1-x}\right)^2$$ on top and bottom, instead. This comes from the difference of cubes formula $$a^3-b^3=(a-b)(a^2+ab+b^2).$$ Had we been dealing with square roots instead of cube roots, then you would use the difference of squares formula, instead, as you are attempting to do implicitly.
A: Remember the identity $$a^3 - b^3 = (a-b)(a^2 + ab + b^2).$$ If you think of $a$ as $(1+x)^{1/3}$ and $b$ as $(1-x)^{1/3}$, you find that multiplying by $$\frac{(1+x)^{2/3} + (1+x)^{1/3}(1-x)^{1/3} + (1-x)^{2/3}}{(1+x)^{2/3} + (1+x)^{1/3}(1-x)^{1/3} + (1-x)^{2/3}}$$ will work.
A: $$
\lim_{x\to 0}\frac{\sqrt[3]{1+x}-\sqrt[3]{1-x}}{x}=\lim_{x\to 0}\frac{\sqrt[3]{1+x}-\sqrt[3]{1-x}}{x}\cdot\frac{\sqrt[3]{(1+x)^2}+\sqrt[3]{1-x^2}+\sqrt[3]{(1-x)^2}}{x(\sqrt[3]{1-x^2}+\sqrt[3]{1-x^2}+\sqrt[3]{(1-x)^2}}=
$$
$$
=\lim_{x\to 0}\frac{1+x-1+x}{x(\sqrt[3]{1-x^2}+\sqrt[3]{(1-x^2}+\sqrt[3]{(1-x)^2}}=\lim_{x\to 0}\frac{2x}{x\left(\sqrt[3]{(1+x)^2}+\sqrt[3]{1-x^2}+\sqrt[3]{(1-x)^2}\right)} 
$$
$$
=\lim_{x\to 0}\frac{2}{\sqrt[3]{(1+x)^2}+\sqrt[3]{1-x^2}+\sqrt[3]{(1-x)^2}}=\lim_{x\to 0}\frac{2}{\sqrt[3]{(1+0)}^2+\sqrt[3]{1-0^2}+\sqrt[3]{(1-0)^2}}=\frac{2}{3}  
$$
A: As $x\to0, |x|<1$ 
Using General Binomial Theorem 
$\displaystyle(1+x)^{\frac13}=1+\frac13x+O(x^2)$ 
and  $\displaystyle(1-x)^{\frac13}=1-\frac13x+O(x^2)$
$$\implies\lim_{x\to0}\frac{\sqrt[3]{1+x}-\sqrt[3]{1-x}}x$$
$$=\lim_{x\to0}\frac{1+\frac13x+O(x^2)-\{1-\frac13x+O(x^2)\}}x$$
$$=\lim_{x\to0}\frac23+O(x)\left(\text{ cancelling }x\text{ as }x\ne0\text{ as }x\to0\right)$$
$$=\cdots$$
