Domain, continuity, boundedness of functions

At what point are the following functions defined? Are they continuous? Are they bounded?

Variant 1
\(\frac{2xy}{x^2 + y^2}\)
Resolution
The function is defined as long as the denominator in the fraction is nonzero, that is as long as \((x,y) \neq (0, 0)\). Likewise, it is continuous at all points not equal to \((0, 0)\), since we are dealing with the ratio of two continuous functions, with the condition that the function in the denominator is nonzero.

We will show that the function is also bounded. For any real numbers \(a, b\), the inequality \(2ab \leq a^2 + b^2\), is valid, since this inequality is equivalent to \(0 \leq (a-b)^2\). Thus \[ \left|\frac{2xy}{x^2 + y^2}\right| = \frac{2|x||y|}{x^2 + y^2} \leq \frac{|x|^2 + |y|^2}{x^2 + y^2} = 1, \] which means that the function is bounded.
Variant 2
\(\cos \frac{1}{xy}\)
Variant 3
\(\frac{1}{1-x^2 - y^2}\)
Variant 4
\( \ln \sqrt{x^2 + y^2} \)
Variant 5
\( \frac{1}{(x-y)^2} \)
Variant 6
\( \frac{ \sin xy}{|x| + |y|} \)
Hint
To show boundedness, we use the inequality \(\sin t \leq |t|\) and we separate the cases \(|x|, |y| \leq 1\) and \(|x| > 1\) or \(|y| > 1\).