What is $\frac{\partial^2(g\circ f)}{\partial x_i \partial x_j}(c)$ If $f:A\to \mathbb R^k$ and $g:B\to \mathbb R$ are two functions of class $C^2$ and their composition is well defined. 
For $c \in A$ what is $$\frac{\partial^2(g\circ f)}{\partial x_i \partial x_j}(c)$$
Is it just $$\frac{\partial^2(g\circ f)}{\partial x_i \partial x_j}(c)=\frac{\partial^2g}{\partial x_i \partial x_j}(f(c))\frac{\partial^2f}{\partial x_i \partial x_j}(c)
$$
If it is how to prove it.
I get from 
$$\frac{\partial^2(g\circ f)}{\partial x_i \partial x_j}(c)=\frac{\partial}{\partial x_i}\frac{\partial(g\circ f)}{ \partial x_j}(c)=\frac{\partial}{\partial x_i}\left( \frac{\partial g}{ \partial x_j}(f(c))\frac{\partial f}{\partial x_j}(c)\right)$$
but not sure how to calculate this, is this a product rule?
 A: We consider open sets  $A\subseteq \mathbb{R}^n$,  $B\subseteq \mathbb{R}^k$ and  $C^2$  functions  $f$ and  $g$
\begin{align*}
&f:A\subseteq      \mathbb{R}^n\to\mathbb{R}^k\\
&g:B\subseteq  f(A)\to\mathbb{R}
\end{align*}
We have real-valued functions
\begin{align*}
&f_1\left(x_1,\ldots,x_n\right),\ldots,f_k\left(x_1,\ldots,x_n\right)\\
&g\left(f_1,\ldots,f_k\right)
\end{align*}
and obtain
\begin{align*}
\frac{\partial\left(g\circ f\right)}{\partial x_j}
&=\frac{\partial g}{\partial  f_1}\,\frac{\partial f_1}{\partial  x_j}
+\frac{\partial g}{\partial  f_2}\,\frac{\partial f_2}{\partial  x_j}+\cdots+\frac{\partial g}{\partial  f_k}\,\frac{\partial f_k}{\partial  x_j}\\
&=\sum_{q=1}^k\frac{\partial g}{\partial f_q}\frac{\partial f_q}{\partial x_j}\tag{1}
\end{align*}
More verbose we can write (1) as
\begin{align*}
\frac{\partial\left(g\circ f\right)}{\partial x_j}\left(x_1,\ldots,x_n\right)
&=\sum_{q=1}^k\frac{\partial g}{\partial f_q}\left(f_1\left(x_1,\ldots,x_n\right),\ldots,f_k\left(x_1,\ldots,x_n\right)\right)\cdot\frac{\partial f_q}{\partial x_j}\left(x_1,\ldots,x_n\right)
\end{align*}

We calculate from (1) the second partial  derivative:
\begin{align*}
\color{blue}{\frac{\partial^2\left(g\circ  f\right)}{\partial  x_i\,\partial x_j}}
&=\frac{\partial  }{\partial x_i}\left(\frac{\partial\left(g\circ f\right)}{\partial x_j}\right)\\
&=\frac{\partial  }{\partial x_i}\left(\sum_{q=1}^k\frac{\partial g}{\partial f_q}\frac{\partial f_q}{\partial x_j}\right)\tag{2}\\
&=\sum_{q=1}^k\frac{\partial  }{\partial x_i}\left(\frac{\partial g}{\partial f_q}\frac{\partial f_q}{\partial x_j}\right)\\
&=\sum_{q=1}^k\left[\left(\frac{\partial }{\partial  x_i}\left(\frac{\partial  g}{\partial  f_q}\right)\right)\frac{\partial  f_q}{\partial x_j}
+\frac{\partial  g}{\partial  f_q}\frac{\partial}{\partial   x_i}\left(\frac{  \partial f_q}{\partial  x_j}\right)\right]\tag{3}\\
&=\sum_{q=1}^k\left(\frac{\partial^2  g}{\partial f_1\partial  f_q}\frac{\partial f_1}{\partial x_i}
+\cdots+\frac{\partial^2  g}{\partial f_k\partial  f_q}\frac{\partial f_k}{\partial x_i}\right)
\frac{\partial  f_q}{\partial x_j}
+\sum_{q=1}^k\frac{\partial g}{\partial f_q}\frac{\partial ^2 f_q}{\partial  x_i\partial x_j}\tag{4}\\
&\,\,\color{blue}{=\sum_{q=1}^k\sum_{r=1}^k\frac{\partial^2 g}{\partial f_r\partial f_q}
\frac{\partial f_r}{\partial x_i}\frac{\partial  f_q}{\partial  x_i}
+\sum_{q=1}^k\frac{\partial g}{\partial f_q}\frac{\partial ^2 f_q}{\partial  x_i\partial x_j}}
\end{align*}

Comment:

*

*In (2) we apply (1).


*In (3) we apply the product  rule for partial derivatives.


*In (4)  we  apply  the chain rule for partial derivatives according to (1).
A: Edit this is what I got from 1 day of thought :D
$f : A \to B, g: B \to \mathbb R^m, A\subseteq \mathbb R^n, B\subseteq \mathbb R^k$
$
\begin{aligned}\frac{\partial^2(g\circ f)}{\partial x_i\partial x_j}(x) 
&=\frac{\partial}{\partial x_i}\left(\frac{\partial(g\circ f)}{\partial x_j}(x)\right) \\&=\frac{\partial}{\partial x_i}\left(\nabla (g\circ f)(x)e_j \right) \\&=\frac{\partial}{\partial x_i}\left(\nabla g(f(x))\nabla f(x)e_j) \right)
\\&=\frac{\partial}{\partial x_i}\left(\nabla g(f(x)) \frac{\partial f}{\partial x_j}(x) \right)\\& 
=\frac{\partial}{\partial x_i}\left(\sum_{p=1}^k  \frac{\partial g}{\partial x_p}(f(x))\frac{\partial f_p}{\partial x_j}(x)\right)
\\&=\sum_{p=1}^k\frac{\partial}{\partial x_i}\left(\frac{\partial g}{\partial x_p}(f(x))\frac{\partial f_p}{\partial x_j}(x) \right)\\
&=\sum_{p=1}^k\left[ \frac{\partial}{\partial x_i}\left(\frac{\partial g}{\partial x_p}(f(x)) \right)\frac{\partial f_p}{\partial x_j}(x)+\frac{\partial g}{\partial x_p}(f(x)) \frac{\partial^2f_p}{\partial x_i\partial x_j}(x) \right]\\
&= \sum_{p=1}^k \frac{\partial}{\partial x_i}\left(\frac{\partial g}{\partial x_p}(f(x)) \right)\frac{\partial f_p}{\partial x_j}(x)+ \sum_{p=1}^k \frac{\partial g}{\partial x_p}(f(x)) \frac{\partial^2f_p}{\partial x_i\partial x_j}(x)\\
&=\sum_{p=1}^k\sum_{l=1}^k\frac{\partial^2g}{\partial x_l\partial x_p}(f(x))\frac{\partial f_l}{\partial x_i}(x)\frac{\partial f_p}{\partial x_j}(x)+\sum_{p=1}^k \frac{\partial g}{\partial x_p}(f(x)) \frac{\partial^2f_p}{\partial x_i\partial x_j}(x)\end{aligned}
$
Also from this we get
$\begin{aligned}D^2(g\circ f)(x)(h,k)&=\sum_{i=1}^n\sum_{j=1}^n\frac{\partial^2(g\circ f)}{\partial x_i\partial x_j}(x) h_ik_j\\
&=\sum_{i=1}^n\sum_{j=1}^n\left(\sum_{p=1}^k\sum_{l=1}^k\frac{\partial^2g}{\partial x_l\partial x_p}(f(x))\frac{\partial f_l}{\partial x_i}(x)\frac{\partial f_p}{\partial x_j}(x)+\sum_{p=1}^k \frac{\partial g}{\partial x_p}(f(x)) \frac{\partial^2f_p}{\partial x_i\partial x_j}(x) \right)h_ik_j\\
\\&=\sum_{i=1}^n\sum_{j=1}^n\left(\sum_{p=1}^k\sum_{l=1}^k\frac{\partial^2g}{\partial x_l\partial x_p}(f(x))\frac{\partial f_l}{\partial x_i}(x)h_i\frac{\partial f_p}{\partial x_j}(x)k_j\right)\\+ \sum_{i=1}^n\sum_{j=1}^n\left(\sum_{p=1}^k \frac{\partial g}{\partial x_p}(f(x)) \frac{\partial^2f_p}{\partial x_i\partial x_j}(x)h_ik_j \right)\\
\\&=\sum_{p=1}^k\sum_{l=1}^k\frac{\partial^2g}{\partial x_l\partial x_p}(f(x))\left(\sum_{l=1}^n\frac{\partial f_l}{\partial x_i}(x)h_i\right)\left(\sum_{p=1}^n\frac{\partial f_p}{\partial x_j}(x)k_j \right)\\
+\sum_{p=1}^k \frac{\partial g}{\partial x_p}(f(x))\left(\sum_{i=1}^n\sum_{j=1}^n \frac{\partial^2f_p}{\partial x_i\partial x_j}(x)h_ik_j\right)\\
&=\sum_{p=1}^k\sum_{l=1}^k\frac{\partial^2g}{\partial x_l\partial x_p}(f(x))(Df_l(x)h)(Df_p(x)k)\\+\sum_{p=1}^k \frac{\partial g}{\partial x_p}(f(x))D^2f_p(x)(h,k)\\&
=\sum_{p=1}^k\sum_{l=1}^k\frac{\partial^2g}{\partial x_l\partial x_p}(f(x))(Df(x)h)_l(Df(x)k)_p\\+\sum_{p=1}^k \frac{\partial g}{\partial x_p}(f(x))\left(D^2f(x)(h,k)\right)_p\\&=D^2g(f(x))(Df(x)h,Dg(x)k)+Dg(f(x))(D^2f(x)(h,k))
\end{aligned}$
That is  we get
$$D^2(g\circ f)(x)(h,k)=D^2g(f(x))(Df(x)h,Dg(x)k)+Dg(f(x))(D^2f(x)(h,k))$$
