A problem with a application of SLLN and CLT and distribution convergence 
Let $Y_{i}\overset{i.i.d}{\sim} N(0,\sigma^{2})$ with $\sigma^{2}$ unknow. I'm trying to prove that $$
    \sqrt{n}\left(\overline{Y}_{n}^{2}- \sigma^2 \right) \overset{D}{\to}  N\left(0,2\sigma^4\right)$$

My approach: I proved that $\overline{Y}_{n}^{2}\overset{D}{\to} \sigma^{2}$ using the strong law large number and using the central limit theorem for random variables i.i.d we have that $\sqrt{n}(\overline{Y}_{n}-\mathbb{E}X_{i})\overset{D}{\to} N(0,\sigma^{2})$, then I was trying use that fact that $f(x)=x^{2}$ and apply the Delta's method, but I can't continue from here because $f(x)=[\mathbb{E}X_{i}]^{2}=0$ and I obtained $\sqrt{n}\overline{Y}_{n}^{2}\overset{D}{\to}N(0,0)$ but its not
desired. 1) How can I solve this problem from here? and can I use that result for conclude that $\overline{Y}_{n}^{2}$ is consistent for $\sigma^{2}$? Note: I proved that $\overline{Y}_{n}^{2}$ is consistent without that result above, but also I'm trying to see if I can conclude the same using the result above.
 A: there are various ways to approach the problem.
The first I get in mind is the more statistical one.
Just to avoid misunderstandings, let's set
$$Y_i \sim N(0;\theta)$$
where $\theta=\sigma^2$
First observe that $T=\overline{Y}_n^2$ is the maximum likelihood estimator for $\sigma^2$ thus is is well known (it is one important property of MLE that you can check in any basic Statistics Textbooks) that
$$\sqrt{I_n(\theta)}\Big(\overline{Y}_n^2-\theta\Big)\dot{\sim}N[0;1]$$
Where $I_n(\theta)$ is the Fischer Information.
$$I_n(\theta)=-nE\Bigg\{\frac{\partial^2}{\partial \theta^2}\log f(x,\theta)\Bigg\}$$

Calculations...

$$f(x,\theta)\propto \theta^{-1/2}e^{-y^2/(2\theta)}$$
$$\log f(x,\theta)=-\frac{1}{2}\log\theta-\frac{y^2}{2\theta}$$
$$\frac{\partial}{\partial\theta}\log f(x,\theta)=-\frac{1}{2\theta}+\frac{y^2}{2\theta^2}$$
$$\frac{\partial^2}{\partial\theta^2}\log f(x,\theta)=\frac{1}{2\theta^2}-\frac{y^2}{\theta^3}$$
$$\mathbb{E}\Bigg[\frac{\partial^2}{\partial\theta^2}\log f(x,\theta)\Bigg]=\frac{1}{2\theta^2}-\frac{\theta}{\theta^3}=-\frac{1}{2\theta^2}$$
$$-n\mathbb{E}\Bigg[\frac{\partial^2}{\partial\theta^2}\log f(x,\theta)\Bigg]=\frac{n}{2\theta^2}=\frac{n}{2\sigma^4}$$

This means that
$$\overline{Y}_n^2\dot{\sim}N\Bigg(\sigma^2;\frac{2\sigma^4}{n}\Bigg)$$
Your are all set!
