Proving a function is continuous using rigorous definition of a limit Im trying to prove that the function $$\begin{cases}f(x,y)=\dfrac{(2x^2y^4-3xy^5+x^6)}{(x^2+y^2)^2}, & (x,y)≠0\\ 0, & (x,y)=0\end{cases}$$ is continuous at point (0,0) using the rigorous defintion of a limit.
Attempting to find the upper limit of the function:
$$|f(x)-f(x_0)|= \left|\frac{(2x^2y^4-3xy^5+x^6)}{(x^2+y^2)^2}-0\right|$$
I see the denominator is always positive so this is equal to
$\dfrac{|2x^2y^4-3xy^5+x^6|}{(x^2+y^2)^2}$.
Using the triangle inequality i know that this is equal or less than
$\dfrac{|(2x^2y^4)-(3xy^5)|+|x^6|}{(x^2+y^2)^2}$.
From here I would like to continue finding expressions which are equal or greater than this, which allow me to cancel some terms against $((x^2+y^2)^2)$.
Im thinking i can write
$$x^6 = (x^2)^3 ≤ (x^2+y^2)^3 $$
for instance, but i am unsure of how to "handle" $|(2x^2y^4)-(3xy^5)$|.
Could someone give me any pointers?

Using the triangle inequality like this:
$$|f(x,y)|\le \frac{|2x^2y^4|+|-3xy^5|+|x^6|}{(x^2+y^2)^2}$$
Notice that
$$\frac{|x^6|}{(x^2+y^2)^2}\le \frac{|x|^6}{x^4}=\frac{x^6}{x^4}=x^2$$
$$\frac{|2x^2y^4|}{(x^2+y^2)^2} \le \frac{2x^2y^4}{y^4}=2x^2$$
$$\frac{|-3xy^5|}{(x^2+y^2)^2} \le \frac{3|x||y|y^4}{y^4}=3|x||y|$$
It is $|f(x,y)| \le 3(x^2+|x||y|)$.
So, noticing that $\sqrt{x^2+y^2}<\delta$ implies $|x|<\delta$ and $|y|<\delta$, you shold be able to conclude with an appropriate choice of $\delta$.
