TITLE: proof continuity of a multivariable function
QUESTION [5 upvotes]: $$
f(x,y) =
 \begin{cases}
 \frac{x^3-xy^2}{x^2+y^2}, & \text{if }(x,y)\not= (0,0) \\
0, & \text{if } (x,y)=(0,0) \end{cases}
$$
how to prove $f(x,y)$ is continuous on $\mathbb{R}^2$? I know it must be continuous on $\mathbb{R}^2$, but according to the definition of continuity, how to use the $\delta- \varepsilon $way to prove?
I know how to prove the continuity at $(0,0)$ but want to know how to prove in the general case other than at $(0,0)$. I can't find a $\delta$ to fit the $\varepsilon$. I mean, let $|f(x,y)-f(x_1,y_1)|<\varepsilon$ 
$(x_1,y_1)$)is any point but not $(0,0)$.
and how to transform it to relate to a $\delta$ in order to have
$|(x,y)-(x_1,y_1)|<\delta$ then $|f(x,y)-f(x_1,y_1)|<\varepsilon$?

REPLY [6 votes]: You can argue by using the following inequality at $(0,0)$:
$$\left|\frac{x^3-xy^2}{x^2+y^2} \right|\leq |x|\frac{x^2}{x^2+y^2}+|x|\frac{y^2}{x^2+y^2}=|x|$$
Therefore, as $(x,y) \to (0,0)$ you have that $x \to 0$ and by the previous inequality $f(x,y) \to 0=f(0,0)$.
In other points than $(0,0)$ the continuity is guaranteed because all the operations made are elementary and well defined.