\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 239/15:}
\begin{description}
\item[a)]
We have $(x^4 + y^2)^2 - 4x^4y^2 = (x^4 - y^2)^2 \geq 0$
for all $(x,y)\in\R^2$.  

To see that $f$ is continuous, it's enough to check continuity at $(0,0)$:
$$
\lim_{(x,y)\to(0,0)} x^2 + y^2 - 2x^2y - 
\frac{4x^6 y^2}{(x^4 + y^2)^2}
= 0 - \lim_{(x,y)\to(0,0)} x^2\frac{4x^4 y^2}{(x^4 + y^2)^2} = 0
$$
since $x^2\to 0$ whereas the magnitude of the other factor is bounded above 
by $1$.  Since $f(0,0) = 0$ by definition, this shows that $f$ is continuous
at $(0,0)$.

\item[b)] A computation shows that 
$$
g_\theta(t) = t^2 - 2t^3 (\cos^2 \theta\sin\theta) - t^4 h(t)
$$
where 
$$
h(t) = 
\frac{4\cos^6\theta\sin^2\theta}{(t^2\cos^4\theta - \sin^2\theta)} 
$$
is defined and $C^\infty$ even at $t=0$ (when $\theta\neq 2n\pi$, we have
$\sin\theta\neq 0$ and this is clear; when $\theta=2n\pi$ the numerator
vanishes altogether and $h(t)\equiv 0$).  Thus 
\begin{eqnarray*}
g'_\theta(t) & = & 2t - 6t^2(\cos^2\theta\sin\theta) - t^3(4h(t) + th'(t)) \\
g''_\theta(t) & = & 2 - 12t(\cos^2\theta\sin\theta) - t^2(12h(t) + 8th'(t) 
+ t^2h''(t))
\end{eqnarray*}
It follows that 
$g'_\theta(0) = 0$, $g''_\theta(0) = 2$.  Thus $g_\theta(t)$ has a strict
local minimum at $t=0$.

However, it doesn't follow that $f$ has a local minimum at $(0,0)$.
In fact, 
$$
f(x,x^2) = x^2 + x^4 - 2x^4 - \frac{4x^{10}}{(2x^4)^2}
         = x^2 - x^4 + x^2 = - x^4,
$$
so that e.g. $\{(1/n,1/n^2)\}$ is a sequence of points converging to $(0,0)$
such that $f(1/n,1/n^2) < 0 = f(0,0)$ for every $n\in\N$.
\end{description}
\end{document}