<s>[INST] jacobian of a dot product of 2 functions take the function $f: \mathbb R^3 \rightarrow \mathbb R^2$ an a function $g: \mathbb R^2 \rightarrow \mathbb R^2 $ defined via $$ f
\begin{pmatrix}
 x\\
y\\
z
\end{pmatrix} =
\begin{pmatrix}
 x+z^3\\
xyz
\end{pmatrix}
$$
and $$ g
\begin{pmatrix}
 s\\
t
\end{pmatrix} =
\begin{pmatrix}
 s^2 + t\\
s+t
\end{pmatrix}
$$
compute the vector  $$ J(g \circ f) \begin{pmatrix} 1\\ 2\\ 3 \end{pmatrix}$$ , that is the jacobian of the dot product
 [/INST] I can see that you have no trouble bar silly mistakes in computing the Jacobian. More precisely, we have :
$$
Df(x,y,z) = \begin{pmatrix}
1 & 0 & 3z^2 \\
zy&xz&xy
\end{pmatrix} ; 
Dg(s,t) = \begin{pmatrix}
2s & 1 \\
1&1
\end{pmatrix}
$$
Now, the chain rule is the following : (with variables $a,b,c$ to avoid confusion)
$$
D(g \circ f)(a,b,c) = Dg(f(a,b,c)) \times Df(a,b,c)
$$
So, in steps , given a point $(a,b,c)$ :


*

*Find $f(a,b,c)$.

*Find $Df$ evaluated at $(a,b,c)$, and $Dg$ evaluated at $\color{green}{f(a,b,c)}$ , and not $\color{red}{(a,b,c)}$.

*Multiply them as matrices, in the order specified in the formula.
For example, you have $(a,b,c) = (1,2,3)$. We find that $f(1,2,3) = (28,6)$.
Now, we have $Df(1,2,3) = \begin{pmatrix}
1 & 0 & 27 \\
6 & 3 & 2
\end{pmatrix}$.
We also have $Dg(28,6) = \begin{pmatrix}
56&1\\1&1
\end{pmatrix}$
And therefore, multiplying them in that order gives 
$$
\begin{pmatrix}
56&1\\1&1
\end{pmatrix} \times \begin{pmatrix}
1 & 0 & 27 \\
6 & 3 & 2
\end{pmatrix} = \begin{pmatrix}
62 & 3 & 1514\\
7 & 3 & 29
\end{pmatrix}
$$
Which is the answer.
</s>