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\vspace*{.3in}
\noindent{\bf Rudin, page 239/6:}
Since $f$ is a rational function whose denominator vanishes only at the
origin, it is clear that the partial derivatives of $f$ exist and are 
continuous everywhere except $(x,y) = (0,0)$.  Now at the origin, we have
$$
D_1 f(0,0) = \lim_{h\to 0} \frac{f(h,0) - f(0,0)}{h} 
           = \lim_{h\to 0} \frac{h\cdot 0 - 0}{h^2 + 0} =
	   = \lim_{h\to 0} \frac{0}{h^2} = 0. 
$$
A similar computation shows that $D_2 f(0,0) = 0$.  

However, if we consider the sequence of points $p_n = (1/n,1/n)$, then
$$
\lim_{n\to\infty} f(p_n) 
= \lim_{n\to\infty} \frac{1/n^2}{2/n^2} = \frac12 \neq 0 = f(0,0) 
= f(\lim_{n\to\infty} p_n).
$$
So $f$ is not continuous at $(0,0)$.
\qed

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