Show that $f$ is $3$-Lipschitz w.r.t the second component Let $$f(x,y) = \frac{xy}{1 + x^2 + y^2}$$
and $$D = \{(x,y) \in \mathbb R^2 \ / \ x^2 + y^2 \le 1\}$$

Show that $\forall$ $(x,y_1), (x,y_2)$ $\in D$, $|f(x,y_1) - f(x,y_2)| \le 3|y_1 - y_2|$

Attempt:
$$|f(x,y_1) - f(x,y_2)| = \ldots = |x|\frac{1 + x^2 + y_1y_2}{(1 + x^2 + y_1^2)(1 + x^2 + y_1^2)}|y_1 - y_2|$$
Then I got stuck. 
I noticed that $(1 + x^2 + y_1^2)(1 + x^2 + y_2^2) = (1 + x^2 + y_1y_2)^2 + (1+x^2)(y_1 - y_2)^2$; but I can't see whether this is useful or not.
Any help? Thanks.

This is elementary using that $|x| \leq 1$ and $1+x^2+y_i^2 \geq 1$ as well as $|y_1y_2| \leq \frac{1}{2}(y_1^2 + y_2^2)$:
$$
\begin{aligned}
\left|\frac{xy_1}{1+x^2+y_1^2} - \frac{xy_2}{1+x^2+y_2^2}\right|
&\leq \left|\frac{y_1}{1+x^2+y_1^2} - \frac{y_2}{1+x^2+y_2^2}\right| \\
&\leq \left|\frac{y_1(1+x^2+y_2^2) - y_2(1+x^2+y_1^2)}{(1+x^2+y_1^2)(1+x^2+y_2^2)}\right| \\
&\leq \left|(y_1-y_2)(1+x^2-y_1y_2)\right| \\
&\leq  (1 + \frac{1}{2}(x^2 + y_1^2) + \frac{1}{2}(x^2 + y_2^2))|y_1-y_2| \\
&\leq 2 |y_1-y_2|
\end{aligned}
$$
