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\vspace*{.3in}
\noindent{\bf Supplementary problem 2: }
The first conclusion of the implicit function theorem tells us that 
$f(x,y) = f(a,b)$ for $(x,y)$ in a neighborhood $W$ of $(a,b)$ 
if and only if $x=g(y)$.  In particular, $f(a,b) = f(a,b)$, so it must
be that $a = g(b)$.

Given then that $g$ is $C^1$, its linear approximation at $b$ has the form
$$
L_g(y) = g(b) + g'(b)(y-b) = a + g'(b)(y-b)
$$
where we compute $g'(b)$ using the Chain Rule as follows.  Let 
$H:V\to\R^{n+m}$ ($V\subset \R^m$ is the neighborhood of $b$ specified in
the theorem) be the function $H(y) = (g(y),y)$.  Then we have
$$
f\circ H(y) = f(g(y),y) = f(a,b)
$$
for all $y\in V$.  So by the Chain Rule, 
$$
f'(H(y))\cdot H'(y) = 0
$$
on $V$---in particular at $y=b$.  Now recall that
$$
f' = (D_x f, D_y f)
$$
where $D_x f:\R^n\to \R^n$ and $D_y f:\R^m\to\R^n$ are the portions of 
$f'$ corresponding to partial derivatives with respect to the $x$ variables and
$y$ variables, respectively.  Moreover, 
$$
H'(y) = \left(\begin{array}{c} g'(y) \\ \id \end{array}\right).
$$
where $g'(y):\R^m\to\R^n$ and $\id:\R^m\to\R^m$.  Plugging these things back
in gives
$$
0 = (D_x f(H(b)),D_y f(H(b)))\cdot
\left(\begin{array}{c} g'(b) \\ \id \end{array}\right)
=
(D_x f(a,b),D_y f(a,b))\cdot
\left(\begin{array}{c} g'(b) \\ \id \end{array}\right)
=
D_x f(a,b)g'(b) + D_y f(a,b).
$$
Solving for $g'(b)$, we arrive at
$$
g'(b) = - D_xf(a,b)^{-1}D_yf(a,b) 
$$
and 
$$
L_g(y) = a - D_xf(a,b)^{-1}D_yf(a,b)(y-b).
$$
\qed