Consider $f:\mathbb{R}^3\to\mathbb{R},(x,y,z)\mapsto x+y+z$ and a differentiable function $g:\mathbb{R}^2\to \mathbb{R}$. What is correctly if I want to apply the chain rule,
1.$D(f(x,y,g(x,y))=Df(x,y,g(x,y))D(\begin{pmatrix} x \\ y\\g(x,y) \end{pmatrix})=(f_x(x,y,g(x,y)),f_y(x,y,g(x,y)),f_{g(x,y)}(x,y,g(x,y))\begin{pmatrix} 1& 0 & 0 \\ 0& 1 & 0\\g_x(x,y)& g_y(x,y) & 0 \end{pmatrix}$
or
2.$D(f(x,y,g(x,y))=Df(x,y,g(x,y))D(\begin{pmatrix} x \\ y\\g(x,y) \end{pmatrix})=(f_x(x,y,g(x,y)),f_y(x,y,g(x,y)),f_{g(x,y)}(x,y,g(x,y))\begin{pmatrix} 1& 0 \\ 0& 1 \\g_x(x,y)& g_y(x,y) \end{pmatrix}$
Sorry, it's my last question about the chain-rule.
It's certainly 2). To see this, note that really we are calculating the derivative of a function $h:\mathbb{R}^2 \rightarrow \mathbb{R}$, where $h(x,y)=f(x,y,g(x,y))$. So we expect the derivative $Dh$ to also be a function from $\mathbb{R}^2$ to $\mathbb{R}$. Your first approach would give a map from $\mathbb{R}^3$, not from $\mathbb{R}^2$.