Prove that $f(x,y) =\frac{x^3-xy^2}{x^2+y^2}$ if $(x,y) \neq (0,0)$ and $f(0,0)=0$, is continuous in $\Bbb R^2$

Question

Prove that $$f(x,y) = \begin{cases} \frac{x^3-xy^2}{x^2+y^2}, & (x,y) \neq (0,0) \\ 0, & (x,y) = (0,0) \end{cases}$$ is continuous on $\Bbb R^2$.

There are more parts to this question, but this is the part that I'm stuck on. I know this should probably be simple, but I just can't wrap my mind around how to show that this function is continuous at (0,0). I can conclude that this function is not differentiable at (0,0), because that's one of the other parts of the problem that I can do.

The solution that I'm looking at effectively just says that f is continuous on $\Bbb R^2$, without any real explanation.

It's my understanding that I could use the $\epsilon-\delta$ definition to show continuity, but is there an easier way?

Also, if there isn't an easier way, I would definitely benefit from seeing the $\epsilon-\delta$ method performed in $\Bbb R^2$. Thanks in advance.

Answer 1

Hint:

$$\left| \frac{x^3-xy^2}{x^2+y^2} \right| \leqslant |x| \frac{x^2}{x^2 + y^2} + |y| \frac{|x||y|}{x^2 + y^2} \leqslant |x| + |y|$$

or @Did's hint which is even more streamlined.