I should show that $\lim \limits_{(x,y)\to(0,0)}2x\log(x^2+y^2)=0$ I should show that
$$\lim_{(x,y)\to(0,0)}2x\log(x^2+y^2)=0$$
which inequality should I use?
 A: You can use polar coordinates as Panda suggested, and that's what I'd normally do, but if you want a more elementary inequality-based argument then you could use $$|x|\le r:=\sqrt {x^2+y^2}$$ where $r$ is just the thing tending to zero.

If you're interested in an inequality-based proof of the limit of $r \log r$ then you can use $r=e^{-u}$ and then this is $$\lim_{u\to\infty} -u/e^u$$ The result follows from the inequality $e^u>1+u+\frac 1 2 u^2$ for $u>0$.
A: Hints:
Since the logarithmic function is monotone ascending and also
$$x>0\,\,\wedge\,\,x^2+y^2\ge x^2\implies 2x\log(x^2+y^2)\ge 2x\log x^2=4x\log x\xrightarrow[x\to 0]{}0$$
for example using l'Hospital. OTOH, by the almost same token
$$x<0\;\wedge\;x^2+y^2\ge x^2\implies 2x\log(x^2+y^2)\le 2x\log x^2=4x\log|x|\xrightarrow [x\to 0]{}0$$
Remember: for $\,x,y\,$ close enough to zero, $\,\log(x^2+y^2)<0\,$ ...
A: Note that since $x^2\le x^2+y^2$, we have $|x|\le\sqrt{x^2+y^2}$. Therefore,
$$
0\le\left|2x\log(x^2+y^2)\right|\le2\sqrt{x^2+y^2}\,\log(x^2+y^2)\tag{1}
$$
Now if we set $u=\sqrt{x^2+y^2}$, we get
$$
\begin{align}
\lim_{u\to0^+}2u\log(u^2)
&=\lim_{u\to0^+}4u\log(u)\\
&=\lim_{v\to\infty}4e^{-v}(-v)\\
&=-4\lim_{v\to\infty}\frac{v}{e^v}\\
&=-4\lim_{v\to\infty}\frac1{e^v}\\[6pt]
&=0\tag{2}
\end{align}
$$
using $u=e^{-v}$ then L'Hospital.
Now, we can use the Squeeze Theorem and $(1)$ and $(2)$ to get that
$$
\lim_{(x,y)\to0}2x\log(x^2+y^2)=0\tag{3}
$$
