Yes/No . Is $f$ bounded in a neighbourhood of $(0,0) ?$

Solution 1:

Yes, $f(x)$ is actually bounded everywhere, not just in a neighborhood of $(0,0)$. Every point that is not $(0,0)$ can take advantage of the argument using $x=r\cos\theta$ and $y=r\sin\theta$, which you have almost correct. Some of the inequalities are in fact equalities, and as I mentioned in the comments, one of your inequalities is only luckily correct:

$$\left|\frac{x^2-y^2}{x^2+y^2}\right| =\left|\frac{r^2\cos^2\theta-r^2\sin^2\theta}{r^2\cos^2\theta+r^2\sin^2\theta}\right| =\left|\frac{r^2(\cos^2\theta-\sin^2\theta)}{r^2(\cos^2\theta+\sin^2\theta)}\right| =|\cos^2\theta-\sin^2\theta|\\\leq|\cos^2\theta|+|\sin^2\theta|\leq1+1=2 $$

Solution 2:

If $(0,0)\ne(x,y)\in\Bbb R^2$ then $$|f(x,y)|=\frac {|x^2+(-y^2)|}{x^2+y^2}\le \frac {|x^2|+|(-y^2)|}{x^2+y^2}=\frac {x^2+y^2}{x^2+y^2}=1.$$