Showing $\lim_{(x,y) \to (0,0)} xy \log(x^2+y^2) = 0$

We have

$$xy \log(x^2+y^2) =(x^2+y^2)\log(x^2+y^2)\cdot \frac{xy}{x^2+y^2}\to 0$$

indeed since $t=x^2+y^2\to 0$

$$(x^2+y^2)\log(x^2+y^2)=t\log t\to 0$$

and since $x^2+y^2\ge 2xy$

$$0\le \left|\frac{xy}{x^2+y^2}\right| \le \frac12$$