Why If $x^2=-\frac{1}{3}$ then $x=\pm\frac{i\sqrt{3}}{3}$ If $x^2=-\frac{1}{3}$ then $x=\pm\frac{i\sqrt{3}}{3}$ according to my text book.
I understand that $x=\pm\sqrt{-\frac{1}{3}}$ but do not fully grasp how to express this as $\pm\frac{i\sqrt{3}}{3}$
I understand that $i$ is $\sqrt{-1}$
As far as I can get with my understanding starting from:
$$x=\pm\sqrt{-\frac{1}{3}}$$
$$x=\pm\frac{\sqrt{-1}}{\sqrt{-3}}$$
$$x=\pm\frac{i}{\sqrt{3}i}$$
$$x=\pm\sqrt{3}$$
Where did I go wrong and how can I arrive at $x=\pm\frac{i\sqrt{3}}{3}$
[Edit]
From the comments I now know that $-\frac{1}{3}$ = $\frac{-1}{3}$
Taking that back into my working I still arrive at a different answer:
$$x=\pm\sqrt{-\frac{1}{3}}$$
$$x=\pm\sqrt{\frac{-1}{3}}$$
$$x=\pm\frac{\sqrt{-1}}{3}$$
$$x=\pm\frac{i}{3}$$
How can I get to $x=\pm\frac{i\sqrt{3}}{3}$

What you did wrong was the fact that
$$-\frac13=\frac{-1}{3}=\frac{1}{-3}$$
and NOT
$$-\frac13=\frac{-1}{-3}$$
After the editing the problem is:
$$\sqrt{-\frac13}=\sqrt{\frac{-1}{3}}=\frac{\sqrt{-1}}{\sqrt{3}}=\frac{i}{\sqrt{3}}=\frac{i}{\sqrt{3}}\frac{\sqrt{3}}{\sqrt{3}}=\frac{i\sqrt{3}}{3}$$
