On finding the asymptotic distribution of the sample variance using the delta method This is an exercise I am stuck on.
Given an IID sample $X_1, \dots, X_n \sim N(\mu, \sigma^2)$ with $\mu \ne 0$ let
$$S_n^2 = \frac{1}{n} \sum_{i = 1}^n ( X_i - \overline{X_n})^2$$
be the sample variance and $\overline{X_n}$ the sample mean. I am tasked in finding the asymptotic distribution of $S_n^2$ using the second order delta method. This says that given a continuous and doubly differentiable function $\phi$ with $\phi'(\theta) = 0$ and an estimator $T_n$ of a parameter $\theta$  such that
$$\sqrt{n} (T_n - \theta) \rightarrow_d T $$ where $T$ is some probability density function then $$\sqrt{n} (\phi (T_n) - \phi(\theta)) \rightarrow_d \frac{1}{2} \phi''(\theta) T^2  $$.
In my exercise I think I have to utilize this method taking as $T_n$ the sample mean, the problem I run into is that I don't know how to define the function $\phi$. I know that
$$ \sum_{i = 1}^n ( X_i - \overline{X_n})^2 = \sum_{i = 1}^n  X_i^2-n \overline{X}_n$$
but this does not help me in defining $\phi$. Moreover as a hint I am given to work with 
$$\frac{S^2_n}{\overline{X_n}}$$
and I can't see why this would help. Any ideas?
EDIT: My theorem was wrong as pointed out by Chaconne in his answer, I correct it for the sake of clarity. I am still searching for the appropriate $\phi$ to utilize to apply the delta method to my exercise.
 A: The second order delta method that you have is incorrect. To see why, take $X_1, 
\dots \stackrel {\text{iid}}\sim \mathcal N(0,1)$ so that
$$
\sqrt n \bar X_n \to_d \mathcal N(0, 1)
$$
and consider $g(z)=z^2$ so $g''(z) = 2$. By the statement as you have it, we should find
$$
\sqrt ng(\bar X_n) \stackrel ? \to_d \frac 12 \left(2\right) Y
$$
where $Y \sim \mathcal N(0, 1)$. But clearly this isn't true as $\sqrt n g(\bar X_n) \geq 0$ always.
The full theorem that you need is rather unsightly at first glance but tells us exactly what to do: 


(source: Jun Shao's Mathematical Statistics)
Let's apply this to our example. We have $k=1$ and $\frac{\partial^2 g}{\partial x^2}(0) \neq 0$ (so we'll have $m=2$) so this simplifies dramatically. In particular, we have
$$
(\sqrt n)^2g(\bar X_n)^2 \to_d \frac 1{2!} \sum_{i_1=1}^1 \sum_{i_2=1}^1 \frac{\partial^2 g}{\partial x_{i_1}\partial x_{i_2}}\big\vert_{x = 0} Y_{i_1} Y_{i_2}.
$$
Now since $Y$ has only one component (i.e. it's a scalar) we have $Y = Y_{i_1} = Y_{i_2}$ so this reduces to
$$
 ng(\bar X_n)^2 \to_d \frac 12 2 Y^2 =_d Y^2.
$$
Let's check this by using our knowledge of the exact distribution of $\bar X_n$. 
We know that $\bar X_n \sim \mathcal N(0, \frac 1n)$ therefore in actuality $$
\left(\sqrt n \bar X_n \right) \sim \mathcal N(0,1) \implies \left(\sqrt n \bar X_n \right)^2 \sim \chi^2_1.
$$
Via the delta method we just showed that
$$
a_n^2 (\bar X_n)^2 = n \bar X_n^2 \to_d \left[\mathcal N(0,1)\right]^2 =_d \chi^2_1.
$$
so we have indeed gotten the correct answer.

Does this give you any directions to try with your particular problem? I'll add more steps in that direction if this doesn't help much.
