Calculating $D(g\circ f)(0,0,0)$ using chain rule

Question

Let $f: \mathbb R^3 \to \mathbb R^2$ satisfy the conditions :
(i) $f(0,0,0) = (1,2)$
(ii) $Df(0,0,0) = \begin{bmatrix} 1 && 2 && 3\\0 && 0&& 0\end{bmatrix}$
Let $g: \mathbb R^2 \to \mathbb R^2$ be defined by the equation
$g(x,y) = (x+2y+1,3xy)$

Find $D (g\circ f)(0,0,0)$

Note 1: This question is taken from a book written by Munkres. $D(g\circ f)(0,0,0)$ means the derivative of $g\circ f$ at $(0,0,0)$.

Note 2: The question has a hint! Use chain rule. So, Can i say that :$D(g\circ f) (0,0,0) = D(g)(f(0,0,0)).D(f)(0,0,0)$
So, The wanted derivative is equal to $D(g)(1,2).\begin{bmatrix} 1 && 2 && 3\\0 && 0&& 0\end{bmatrix}$

And since $D(g)(1,2)=g(1,3xy)+g(x+2y+1,2)=(2+6xy,9xy)+(x+2y+6,6x+12y+6)=(6xy+x+2y+8,9xy+6x+12y+6)$, How should i multiplicate $D(g)(1,2)$ into $\begin{bmatrix} 1 && 2 && 3\\0 && 0&& 0\end{bmatrix}$? (If my way of solving the problem is correct)

Answer 1

$$Dg(x,y) = \begin{pmatrix} 1 & 2\\ 3y & 3x \end{pmatrix}$$ So $$Dg(1,2) = \begin{pmatrix} 1 & 2\\ 6 & 3 \end{pmatrix}$$

And like you said, $$D(g\circ f)(0,0,0) = Dg(f(0,0,0))\cdot Df(0,0,0) = \begin{pmatrix} 1 & 2\\ 6 & 3 \end{pmatrix} \cdot \begin{pmatrix} 1 & 2 & 3 \\ 0 & 0 & 0 \end{pmatrix}$$ $$D(g\circ f)(0,0,0) = \begin{pmatrix} 1 & 2 & 3 \\ 6 & 12 & 18 \end{pmatrix}$$