Showing that the function $f(x,y)=x\sin y+y\cos x$ is Lipschitz I wanted to show that  $f(x,y)=x\sin y+y\cos x$ sastisfy Lipschitz conditions. but I can't separate it to $L|y_1-y_2|$. 

According to my lecturer, the Lipschitz condition should be $$|f(x,y_1)-f(x,y_2)|\le  L|y2-y1|$$ 
I was able show that   $x^2+y^2$ in the rectangle $|x|\le a$, $|y|\le b$  satisfies the Lipschitz condition, with my $L=2b$. But I had problem showing this for $f(x,y)=x\sin y+y\cos x $.

Since $\frac {\partial f} {\partial y} =x \cos y+ \cos x$, it follows that
$$\left| \frac {\partial f} {\partial y} \right| = |x \cos y+ \cos x| \le |x \cos y|+|\cos x| = |x| \, |\cos y|+|\cos x| \le a+1$$
since $|x| \le a$, $|\cos y| \le 1$ and $|\sin y| \le 1$. This means that $|f(x,y_1) - f(x,y_2)| \le (a+1) \, |y_1 - y_2|$.
