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\noindent{\bf Rudin, page 239/8:}
Since $f$ is differentiable at $x$, all partial derivatives $Df_1,\dots Df_n$
exist at $x$ and $Df(x) = (Df_1(x),\dots,Df_n(x))$.  And if $f$ has a local 
maximum at the point $x = (x_1,\dots, x_n)$, then for any $1\leq k \leq n$, 
the one-variable function 
$$
g_k(t) := f(x_1,\dots,x_{k-1},t,x_{k+1},\dots, x_n)
$$ 
has a local maximum
at $t = x_k$.  In particular $g_k'(x_k) = 0$.  But $g_k'(x_k)$ is just 
$D_k f(x)$.  Hence $Df(x) = (0,\dots,0)$.
\qed
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