= Solution
The <partial derivative>[partial derivatives] satisfy a global bound of $1$ in absolute value. For example,
$$
|\partial_xf(x,y)|\leq\frac{2|x|}{(1+x^2)^2}\leq1,
$$
because $2|x|\leq1+x^2$ and $1+x^2\geq1$; the same argument applies to $\partial_y f$. Apply the one-dimensional <mean value theorem> first along a horizontal segment and then along a vertical segment:
$$
\begin{aligned}
|f(x,y)-f(u,v)|
&\leq|f(x,y)-f(u,y)|+|f(u,y)-f(u,v)|\\
&\leq\boxed{|x-u|+|y-v|}.
\end{aligned}
$$
Thus $f$ is <Lipschitz continuity>[Lipschitz] with constant $1$ for the coordinate-sum <norm>.
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