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\begin{document}
\vspace*{.3in}
\noindent{\bf Rudin, page 239/14:}
\begin{description}
\item[a)]
For $(x,y) \neq (0,0)$, a quick computation shows that
$$
D_1 f(x,y) = \frac{x^4 + 3x^2 y^2}{(x^2 + y^2)^2}, \quad
D_2 f(x,y) = \frac{-2x^3 y}{(x^2+y^2)^2}.
$$
Taking $D_1(f(x,y))$, for example, we note that both $x^4$ and $x^2y^2$ are
smaller than $|(x,y)|^4$.  Hence
$$
|D_1 f(x,y)| \leq \frac{4|(x,y)|^2}{|(x,y)|^2}  = 4
$$
for all $(x,y)\in\R^2-(0,0)$.  Likewise,
$$
|D_2 f(x,y)| \leq \frac{2|(x,y)|^4}{|(x,y)|^4} = 2.
$$
Finally, for $(x,y) = (0,0)$ one computes
$$
D_1 f(0,0) = \lim_{h\to 0} \frac{(h - 0)}{h} = 1
$$
and, in the same fashion, $D_2 f(0,0) = 0$.

\item[b)]
Let us write $\V u = (s,t)$.  Then
\begin{eqnarray*}
D_{\V u} f(0,0) & = & \lim_{h\to 0} \frac{f(h\V u) - f(\V 0)}{h} \\
                & = & \lim_{h\to 0} \frac{h^3s^3/(h^2s^2 + h^2t^2)}{h}
		= \frac{s^3}{s^2 + t^2} = s^3,
\end{eqnarray*}
since $\V u$ is a \emph{unit} vector.  This shows that $D_{\V u}f(0,0)$ exists 
and, moreover, since $|s|\leq |\V u| \leq 1$, we have 
$|D_{\V u}f(0,0)| \leq 1$.

\item[d)]
Were $f$ actually differentiable at $(0,0)$, $D_{\V u} f$
would be linear in $\V u$.  But from the formula computed in part b), we see
that 
$$
D_{(0,1)} f(0,0) + D_{(1,0)} f(0,0) = 0 + 1 \neq \frac12 = D_{(1,1)}
f(0,0).
$$



\end{description}
\end{document}