\documentclass[11pt]{article}
\usepackage{graphicx}
\usepackage{amssymb}
\usepackage{epstopdf}
\DeclareGraphicsRule{.tif}{png}{.png}{`convert #1 `dirname #1`/`basename #1 .tif`.png}

\textwidth = 6.5 in
\textheight = 9 in
\oddsidemargin = 0.0 in
\evensidemargin = 0.0 in
\topmargin = 0.0 in
\headheight = 0.0 in
\headsep = 0.0 in
\parskip = 0.2in
\parindent = 0.0in

\newtheorem{theorem}{Theorem}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{definition}{Definition}

\begin{document}
\vspace*{.3in}
\noindent{\bf Rudin, page 239/23:}
The linear approximation of $f(x,y_1,y_2)$ about the point $(0,1,-1)$ is 
$$
L(x,y_1,y_2) = f(0,1,-1) + f'(0,1,-1)
\left(\begin{array}{c} x \\ y_1-1 \\ y_2 +1 \end{array}\right)
= 
\left(\begin{array}{c} 0 \\ 0 \\ 0 \end{array}\right)
+ 
\left(\begin{array}{ccc} 1 & 0 & 1 \end{array}\right)
\left(\begin{array}{c} x \\ y_1-1 \\ y_2+1 \end{array}\right)
= 
x + y_2 - 1.
$$
One way to restate the implicit function theorem is to say that you
can locally solve $f = 0$ for some of the variables in terms of the others
if you can do the same for the linear approximation.  So that's we do
here---we want to solve $f(x,y_1,y_2) = 0$ for $x$ in terms of
$y_1$ and $y_2$, so we replace $f$ by $L$ and get
$$
x + y_2 - 1 = 0 \Rightarrow x = 1 - y_2,
$$ 
where the right side is the linear approximation of the implicit function
$x = g(y_1,y_2)$.  In particular, 
$$
\frac{\partial g}{\partial y_1}(1,-1) = 0, \quad
\frac{\partial g}{\partial y_2}(1,-1) = -1.
$$ 
\end{document}