<s>[INST] Is the function $\frac{y(x^4+4x^2y^2-y^4)}{(x^2+y^2)^2}$ Continuous at $(0,0)$ The function is defined piece-wise with $f(0,0) = 0$, so I took the limit $(x,y) \rightarrow (0,0)$
I proceeded like this:
$$\frac{y(x^4+4x^2y^2-y^4)}{(x^2+y^2)^2}  \leq \frac{|y|(x^4+4x^2y^2-y^4)}{(x^2+y^2)^2} \leq \frac{|y|(x^4+4x^2y^2+y^4)}{(x^2+y^2)^2} 
$$
$$= \frac{|y|(x^2+y^2)^2+|y|2x^2y^2}{(x^2+y^2)^2} =  |y| + \frac{|y|2(xy)^2}{(x^2+y^2)^2} \leq |y| + \frac{|y|2(\frac{x^2+y^2}{2})^2}{(x^2+y^2)^2} = |y| + \frac{|y|}{2} \rightarrow0$$ 
Hence by the sandwich/squeeze theorem, the original function also goes to $0$
Is this answer ok?
 [/INST] Your proof is correct but here is shorter way.
Since $|x|,|y|\le \|(x,y)\|$
we get  $$\left|\frac{y(x^4+4x^2y^2-y^4)}{(x^2+y^2)^2}\right|=\left|\frac{y(x^4+4x^2y^2-y^4)}{\|(x,y)\|^4}\right|\le 6\frac{\|(x,y)\|^5}{\|(x,y)\|^4} =\|(x,y)\|\to 0$$
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