I was trying to get the asymptotic distribution of sample variance using multivariate delta method and without normality assumption.
So I defined the random vector $ z = \left( \begin{matrix} X \\ X^2 \\ \end{matrix} \right) $ and the function $g$ which acts on components : $$g(z_1,z_2)=z_2 - z_1^2 $$ Clearly $g(E[z])=\sigma_X^2$.
Taking a random sample, the random vector: $z_n= \left( \begin{matrix} \frac{1}{n}\sum_i X_i \\ \frac{1}{n}\sum_i X_i^2 \\ \end{matrix} \right) $ a plugin estimator for $\sigma_X^2$ is $g(z_n)=S_X^2 $ (sample variance).
Knowing that $z_n$ converges asymptotically to a normal distribution thanks to central limit theorem: $${{z}_{n}}\xrightarrow{d}N\left( E[z],\frac{1}{n}\left( \begin{matrix} V[X] & \operatorname{cov}(X,{{X}^{2}}) \\ \operatorname{cov}({{X}^{2}},X) & V[{{X}^{2}}] \\ \end{matrix} \right) \right) $$ we can apply multivariate delta method.
I get stuck in the estimation of variance by delta method which should be $$({{\left. \nabla g \right|}_{E[X]}})^t \left( \begin{matrix} V[X] & \operatorname{cov}(X,{{X}^{2}}) \\ \operatorname{cov}({{X}^{2}},X) & V[{{X}^{2}}] \\ \end{matrix} \right){{\left. \nabla g \right|}_{E[X]}} $$ with ${{\left. \nabla g \right|}_{E[X]}} = \left( \begin{matrix} -2E[X] \\ 1 \\ \end{matrix} \right) $ .
Performing the matrix product as above I end up with an asymptotic variance which does not match the one I found on textbooks , which is: $$V[(X-E[X])^2]$$ What I'm missing?