Proving continuity at $0$ for $f(x,y)=\dfrac{x^2y}{x^2+y^2}$ Here is my attempt at proving the problem below.  
$$\bigg|\dfrac{x^2y}{x^2+y^2}\bigg| \leq \bigg|\dfrac{x^2y}{2xy}\bigg| \text{ by the hint}$$ 
$$=\bigg|\dfrac{x}{2}\bigg| \leq |x| $$
Since we're using the delta epsillon definition to prove continuity, it should then follow that $|x|\leq|(x,y)|$.  Have I missed anything or done something incorrectly here?
Thanks! 


 A: Yes, it is correct. $f(x,y)=0$ if $(x,y) \neq (0,0)$, so in order to prove the continuity at $(0,0)$ you have to prove that
$$\lim_{(x,y)\to(0,0)}\frac{x^2y}{x^2 + y^2} =0$$
By the $AM-GM$ inequality we have that
$$\frac{x_1+...+x_n}{n} \geq (x_1 \cdot ... \cdot x_n)^{1/n}$$
So taking $x_1 =x^2 , x_2=y^2$:
$$\frac{x^2+y^2}{2} \geq (x^2 y^2)^{1/2} \implies x^2 + y^2 \geq 2 |xy| $$ 
And as you have proved
$$ \bigg|\dfrac{x^2y}{x^2+y^2}\bigg| \leq |x|$$
The absolute value of the component of a vector is always less or equal than the norm of the vector, so
$$\bigg|\dfrac{x^2y}{x^2+y^2}\bigg| \leq |x| \leq \|(x,y)\|$$
Now let $\epsilon \gt0$. Taking $\delta = \epsilon$:
$$\text{if $(x,y)$ is such that } 0 \lt \|(x,y)-(0,0) \| \lt \delta \implies  \bigg|\dfrac{x^2y}{x^2+y^2} - 0 \ \bigg| \lt \epsilon$$
A: Since $|x|\to 0$ when $|(x,y)|\to 0$, we have the result. In this case, you don't need the hint. Just use
$$
\left|\frac{x^2}{x^2+y^2}\right| \leq 1
$$
so
$$
\left|\frac{x^2y}{x^2+y^2}\right| \leq 1\cdot |y| \to 0 \qquad \text{if $|(x,y)|\to 0$}
$$
