I would like to know if I am wrong with the following:
Prove or disprove the continuity of the function $$f(x, y) = \begin{cases} \frac{4x^3y^3(y^4-x^8)}{(x^8+y^4)^2} & (x, y) \neq (0, 0) \\\\ 0 & (x, y) = (0, 0) \end{cases}$$ at the origin.
Here is what I did:
\begin{equation} \begin{split} \left|\dfrac{4x^3y^3(y^4-x^8)}{(x^8+y^4)^2} \right| & \leq \frac{4(y^4-x^8) x^2y^2 \vert x\vert\vert y \vert}{(x^8 + y^4)^2} \\\\ & \leq \frac{4(y^4+x^8) x^2y^2 \vert x\vert\vert y \vert}{(x^8 + y^4)^2} \\\\ & \leq \frac{4x^2y^2|x||y|}{x^8 + y^4} \\\\ & \leq \frac{4x^2y^2 |x||y|}{2x^4y^2} \\\\ & \leq \frac{2(|x|+|y|)(|x| + |y|)}{x^2} \\\\ & \leq \frac{2C^2(x^2+y^2)}{x^2} \end{split} \end{equation}
Where I clarify that:
Second line: I majorized $y^4-x^8 \leq y^4+x^8$ because since $x, y$ both go to zero, then adding is always greater than substracting
Line three to line four: AM/GM: $x^8+y^4 \geq 2\sqrt{x^8y^4} = 2x^4y^2$ and then I reversed
Line four to line five: $|x| \leq |x| + |y|$ and repeat for $|y|$.
Last line: the $\ell^2$ norm: $|x| + |y| \leq C \sqrt{x^2+y^2}$ twice (the constant $C$ is positive and non zero, and it's not important here)
At this point I'm not sure how to conclude, because either there is no possibility to compress this in terms of a distance that goes to zero, hence no continuity, or there is some magic tool I do not know.
NOTE: No polar coordinates. Only cartesian and majorizations.