Using Squeeze Theorem to compute $\lim_{(x,y)\to(0,0)} \frac{x²y}{x^2+xy+y^2} = 0$ Can you help me to show that $\lim_{(x,y)\to(0,0)} \frac{x²y}{x^2+xy+y^2} = 0$ with the squeeze theorem ?
I'm running out of ideas to bound $|\frac{x²y}{x^2+xy+y^2}|$. I was thinking  to use $|xy|\leq |x^2+xy+y^2|$, which seems to be a right inequality but don't know to to show it.

$2xy=(x+y)^2-x^2-y^2,$ so $$|f(x,y)|=\frac{x^2|y|}{|x^2+xy+y^2|}=\frac{2x^2|y|}{(x+y)^2 + x^2 + y^2} \le \frac{2x^2|y|}{x^2+y^2}= |x|\frac{2|xy|}{x^2+y^2} \le |x| \frac{x^2+y^2}{x^2+y^2}=|x|$$
where we also used $2|xy| \le x^2+y^2$ which follows from $(|x|-|y|)^2\ge 0$. Sending $(x,y)\to 0$ gives the result.
A now deleted comment informed me that there is an shorter proof:
$$ |f(x,y)|=\frac{x^2|y|}{|x^2+xy+y^2|}=\frac{2x^2|y|}{(x+y)^2 + x^2 + y^2}  \le 2|y|.$$
This follows because $x^2 \le (x+y)^2 + x^2 + y^2$.  Together with the above, we have the improved bound
$$ |f(x,y)| \le \min(|x|,2|y|).$$
