Answer
The limit is
$$\lim_{(x,y)\to(0,0)}(x^2+y^2)\ln(x^2+y^2) = 0.$$
Work Step by Step
We have
$$\lim_{(x,y)\to(0,0)}(x^2+y^2)\ln(x^2+y^2).$$
Introduce the polar coordinates:
$$\rho^2=x^2+y^2,\quad \tan\phi=\frac{y}{x},\\ \quad (x,y)\to0\Rightarrow \rho\to 0.$$
Now we have
$$\lim_{(x,y)\to(0,0)}(x^2+y^2)\ln(x^2+y^2)=\lim_{\rho\to0}\rho^2\ln\rho^2 = \lim_{\rho\to0}\frac{\ln\rho^2}{\frac{1}{\rho^2}}.$$
The last transformation is performed so that we can apply L'Hopital's rule. Namely, when $\rho\to0$ then $\ln\rho^2 \to -\infty $ (because when the argumet of $\ln$ tends to zero its value tends to $-\infty$) and the denominator $1/\rho^2$ tends to $\infty$. Since both the numerator and the denominator tend to some kind of infinity the L'Hopital's rule is applicable:
$$\lim_{\rho\to0}\frac{\ln\rho^2}{\frac{1}{\rho^2}} = \lim_{\rho\to0}\frac{(\ln\rho^2)'}{\left(\frac{1}{\rho^2}\right)'}=\lim_{\rho\to0}\frac{2\rho\frac{1}{\rho}}{-\frac{2}{\rho^3}}=\\\lim_{\rho\to0}-\rho^3 = 0.$$
where in the last step we used the method of substitution.
