Exercise in "Analysis on Manifolds" - can this be done in a more elegant way? This is about exercise 4(c) on page 79 of Analysis on Manifolds by
Munkres.  I think I have solved the exercise, but my solution seems horribly
complicated and I'm wondering if I'm missing something obvious that can make
this more elegant.
We start with a function $F:\mathbb{R}^2\to\mathbb{R}$ of class
$\mathcal{C}^2$ with $F(0,0)=0$ and $DF(0,0)=\bigl[2\;3\bigr]$.
We define $G:\mathbb{R}^3\to\mathbb{R}$ by
$$ G(x,y,z) = F(x+2y+3z-1,x^3+y^2-z^2) $$
noting that for $\mathbf{a}=(-2,3,-1)$ we have $G(\mathbf{a})=F(0,0)=0$.
In part (a) you have to prove that you can find a neighborhood $B$ of $(-2,3)$
and a function $g:B\to\mathbb{R}$ such that $G(x,y,g(x,y))=0$ on $B$ and $g(-2,3)=-1$.
This is just the implicit function theorem (which this section of the book is
about) and it also tells us that $g$, too, is $\mathcal{C}^2$.
Part (b) asks you to compute $Dg(-2,3)$.  This is straightforward and my result is
$\bigl[-19/6 \; -11/6\bigr]$.
Now comes part (c) which is marked as "hard":  Given $D_1D_1F=3$, $D_1D_2F=-1$
and $D_2D_2F=5$ at $(0,0)$, compute $D_2D_1g(-2,3)$.
My solution goes like this:  We know that
$$\begin{aligned}
  &\bigl[D_1G(x,y,g(x,y)) \;\; D_2G(x,y,g(x,y))\bigr] \\
  + \; &\bigl[D_3G(x,y,g(x,y))\bigr] \cdot \bigl[D_1g(x,y) \;\; D_2g(x,y)\bigr]
\end{aligned}$$
is always $\mathbf{0}$ for $(x,y)\in B$ (from theorem 9.1 in the book).
We call the first component of this term $L(x,y)$:
$$L(x,y) = D_1G(x,y,g(x,y)) + D_3G(x,y,g(x,y)) \cdot D_1g(x,y)$$
This gives us a function of class $\mathcal{C}^1$ on $B$ and because it is
constant, its derivative must vanish.
Now comes the heavy lifting - we compute $D_2L(x,y)$ which looks like this:
$$\begin{aligned}
&2 D_2D_1F(I^\ast) \bigl[y-g(x,y) D_2g(x,y)\bigr]  \\
+\;&D_2D_1g(x,y) \bigl[3 D_1F(I^\ast)-2 g(x,y) D_2F(I^\ast)\bigr]
+ D_1D_1F(I^\ast) \bigl[3 D_2g(x,y)+2\bigr] \\
+\;&D_1g(x,y) \Bigl[-2 D_2F(I^\ast) D_2g(x,y) +6 D_2D_1F(I^\ast) \bigl[y-g(x,y) D_2g(x,y)\bigr] \\
  &\phantom{D_1g(x,y)}\,-2 g(x,y) \Bigl(2 D_2D_2F(I^\ast) \bigl[y-g(x,y) D_2g(x,y)\bigr] \\
  &\phantom{D_1g(x,y)-2 g(x,y)} \;     +D_2D_1F(I^\ast) \bigl[3 D_2g(x,y)+2\bigr] \Bigr)\\
  &\phantom{D_1g(x,y)}\,+3 D_1D_1F(I^\ast)\bigl[3 D_2g(x,y)+2\bigr] \Bigr]\\
+\;&3 x^2 \Bigl(2 D_2D_2F(I^\ast)\bigl[y-g(x,y) D_2g(x,y)\bigr] +D_2D_1F(I^\ast)\bigl[3 D_2g(x,y)+2\bigr] \Bigr)
\end{aligned}$$
(I've used $I^\ast$ as an abbreviation for $I(x,y,g(x,y))$
with $I(x,y,z)$ meaning $(x+2y+3z-1,x^3+y^2-z^2)$, otherwise the term would be even longer.)
If we now substitute $(-2,3)$ for $(x,y)$ and use all the values we know
(including the fact that $D_2D_1F$ and $D_1D_2F$ must be identical on $B$), we
get, after a long computation,
$$ 0 = D_2L(-2,3) = 5767/36 + 12 D_2D_1g(-2,3) $$
which eventually leads to: $D_2D_1g(-2,3) = -5767/432$.
I somehow can't believe that Munkres intended his students to go through this
huge computation.  (And, of course, I'm not even sure how many mistakes I've
made.  I used Mathematica, but the book is from 1991.)  Any thoughts?
 A: The last part surely implies some calculations, but one can take a systematic approach to make them comprehensible. I will use a short notation ($F_i = D_iF$, $F_{ij}=D_iD_jF$) and omit the proofs of differentiability.
In $B$ we have: 
$$
  G = 0,
$$
$$
D_1[G(\dots)] =  G_1+G_3g_1=0,
$$
$$
D_2[G_1(\dots)+G_3(\dots)g_1] = (G_{21}+G_{31}g_2)+\big(G_{23}+G_{33}g_2\big)g_1+G_3g_{21}=0
$$
In matrix form:
$$
-G_3g_{21} = \begin{pmatrix}1\\g_1\end{pmatrix}^T
\begin{pmatrix}G_{21}&G_{31}\\G_{23}&G_{33}\end{pmatrix}
\begin{pmatrix}1\\g_2\end{pmatrix}
=
\frac{1}{36}\begin{pmatrix}6\\-19\end{pmatrix}^T
\begin{pmatrix}G_{21}&G_{31}\\G_{23}&G_{33}\end{pmatrix}
\begin{pmatrix}6\\-11\end{pmatrix}
$$
Now the trick is how to caculate $G_{ij}$ relatively easy. Let's denote $u^1=x+2y+3z$ and $u^2 = x^3+y^2-z^2$
Since $G_i = \sum_{\alpha=1,2}F_\alpha u^\alpha_i$,
$$
G_{ij} = \sum_{\alpha=1,2}F_\alpha u^\alpha_{ij} + \sum_{\alpha=1,2}\sum_{\beta=1,2}F_{\alpha\beta} u^\alpha_{i}u^\alpha_{j}.
$$
Function $u^1$ is linear, so all second derivatives $u^1_{ij}=0$, and first term is just $F_2u^2_{ij}$, which is diagonal matrix and is non-zero only for $G_{33}$:
For second term $S_{ij}$ not much can be done, but we can rewrite it as a bilinear form again:
$$
S_{ij} = v_i^T \begin{pmatrix}3&-1\\-1&5\end{pmatrix}v_j,
$$
where $v_i = [u^1_i, u^2_i]$: $v_1 = [1, 12]$, $v_2 = [2, 6]$, $v_3=[3, 2]$.
We have:
$$
\begin{pmatrix}G_{21}&G_{31}\\G_{23}&G_{33}\end{pmatrix} = 
\begin{pmatrix}S_{21}&S_{31}\\S_{23}&S_{33}-2F_2\end{pmatrix}=
\begin{pmatrix}336&91\\56&29\end{pmatrix}
$$
Finally, $G_3=3F_1+2F_2=12$: 
$$
-12g_{21}=\frac1{36}\times 5767
$$
