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.
 A: Let us use the mean value theorem for functions with two variables, i.e., $f(x,y)-f(x,z)=\frac{\partial{}f}{\partial{}y}\Big|_{y=\xi}(y-z)$,
where $\xi$ is between $y$ and $z$.
Define
$f(x,y):=\frac{xy}{1+x^{2}+y^{2}}$ for $(x,y)\in{}D:=\{(x,y):\ x^{2}+y^{2}\leq1\}$.
It should be noted that $(x,y)\in{}D$ implies $|x^{2}-y^{2}|\leq{}x^{2}+y^{2}\leq1$ and $|x|\leq\sqrt{x^{2}+y^{2}}\leq1$. Then, we can compute that
$f(x,y)-f(x,z)=\frac{x(1+x^{2}-\xi^{2})}{(1+x^{2}+\xi^{2})^{2}}(y-z)$,
which yields
$|f(x,y)-f(x,z)|\leq\frac{|x|(1+x^{2}+\xi^{2})}{(1+x^{2}+\xi^{2})^{2}}|y-z|\leq\frac{1(1+1)}{(1+0)^{2}}|y-z|=2|y-z|$
for all $(x,y)\in{}D$, where $\xi$ is some number between $y$ and $z$.
A: From where you have it, you're almost there.  note that $x,y_1,y_2\in D$,so they all have absolute value at most 1.  Then we have $0\le x^2\le 1$, and $-1\le y_1y_2\le 1$,  hence in the numerator of your big fraction, we have $0\le 1+x^2+y_1y_2\le 3$.  The denominator is at least $1$, hence the fraction is bounded between $0 and 3$.   $|x|$ is likewise bounded between 0 and 1, so the coefficent is bounded between $0$ and $3$.   So we are good as long as we can take care of that 0 case.  The only way for it to be 0 is for $y_1y_2=-1$ and $x=0$, which means we have $(0,1)$ and $(0,-1)$.  Since f at both those points is 0, we are done.
A: This is elementary using that $|x| \leq 1$ and $1+x^2+y_i^2 \geq 1$ as well as $|y_1y_2| \leq \frac{1}{2}(y_1^2 + y_2^2)$:
$$
\begin{aligned}
\left|\frac{xy_1}{1+x^2+y_1^2} - \frac{xy_2}{1+x^2+y_2^2}\right|
&\leq \left|\frac{y_1}{1+x^2+y_1^2} - \frac{y_2}{1+x^2+y_2^2}\right| \\
&\leq \left|\frac{y_1(1+x^2+y_2^2) - y_2(1+x^2+y_1^2)}{(1+x^2+y_1^2)(1+x^2+y_2^2)}\right| \\
&\leq \left|(y_1-y_2)(1+x^2-y_1y_2)\right| \\
&\leq  (1 + \frac{1}{2}(x^2 + y_1^2) + \frac{1}{2}(x^2 + y_2^2))|y_1-y_2| \\
&\leq 2 |y_1-y_2|
\end{aligned}
$$
