<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] Yes your proof looks good, very good.
Your problem is a special case of this result: Suppose $p(x,y),q(x,y)$ are homogeneous polynomials with $\deg p > \deg q,$ and $q(x,y)\ne 0$ except at $(0,0).$ Then
$$\lim_{(x,y)\to (0,0)}\frac{p(x,y)}{q(x,y)}=0.$$
A popular way of proving this is to go to polar coordinates.
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