From wiki, I found that the natural method of moments estimator of Skewness:
$$b_{1}={\frac{m_{3}}{s^{3}}}={\frac{\frac1n\sum_{i=1}^n(x_i-\overline {x})^3}{\sqrt{{\frac1{n-1}}\sum_{i=1}^{n}(x_i-\overline{x})^2}^3}}={\frac{{\frac{1}{n}}\sum_{i=1}^{n}(x_i-\overline{x})^3}{\left[\frac1{n-1}\sum_{i=1}^{n}(x_i-\overline{x})^2\right]^{3/2}}}$$
Along with another estimator (maybe unbiased):
$$G_1=\frac{k_3}{k_2^{3/2}}=\frac{n^{2}}{(n-1)(n-2)}\frac{m_3}{s^3}=\frac{\sqrt{n(n-1)}}{n-2}\left[\frac{\frac1n\sum_{i=1}^n(x_i-\bar{x})^3}{\left(\frac1n\sum_{i=1}^{n}{{\left(x_i-\bar{x}\right)}^2}\right)^{3/2}}\right]$$
Then it says that under the assumption of $X$ normal $\sqrt {n}b_{1}\xrightarrow{d}N(0,6)$ which indicate that $\operatorname{Var}(b_1)=\frac{6}{n}$
Furthermore, it also gives out the variance of $\operatorname{Var}(G_{1})={\dfrac {6n(n-1)}{(n-2)(n+1)(n+3)}}$
Now I would have the following 2 questions:
1, Can anyone help to provide the proof of $\operatorname{Var}(b_1)=\frac{6}{n}$
2, As far as I know $\operatorname{Var}(Cx)=C^2Var(x)$ this indicate that $\operatorname{Var}(G_1)=\operatorname{Var}\left(\dfrac{n^{2}}{(n-1)(n-2)}b_1\right)=\dfrac{6n^3}{(n-1)^2(n-2)^2}$. Why the variance on wiki different.
You don't have $\def\Var{\operatorname{Var}}\Var(b_1)=\frac6n$, only that $\Var(b_1)\sim\dfrac6n$ (i.e. $n\Var(b_1)/6\to 1$ as $n\to\infty$) coming from the weak convergence $\sqrt{n}b_1\xrightarrow{d}N(0,6)$.