<s>[INST] Calculating a multivariable limit. Find if it exists the limit : $$\lim_{(x,y)\to(0,0)}\frac{x^4y^4}{(x^2+y^4)^3}$$
I've tried the following :
1st attempt : Using polar coordinates:
Set $x= r\cos\theta$ and $y=r\sin\theta$
$$\frac{x^4y^4}{(x^2+y^4)^3}=\frac{r^8\cos^4\theta \sin^4\theta}{(r^2\cos^2\theta+r^4\sin^4\theta)^3}=\frac{r^2\cos^4\theta\sin^4\theta}{(\cos^2\theta+r^2\sin^4\theta)^3}$$
$$\lim_{r\to0}\frac{r^2\cos^4\theta\sin^4\theta}{(\cos^2\theta+r^2\sin^4\theta)^3}=\frac{0}{\cos^6\theta}$$
Now the limit mentioned above would be equal to zero if and only if the denominator is different than zero. 
Hence we need to calculate the limit in the case where $\theta = \frac{\pi}{2}+k\pi$ and compare it with the precalculated limit.
However I was not able to get rid of the indeterminate form.
2nd attempt : Choosing a specific path $y = ax$.
$$\frac{x^4y^4}{(x^2+y^4)^3}=\frac{a^4x^2}{(1+a^4x^2)^3}$$
$$\lim_{x\to0}\frac{a^4x^2}{(1+a^4x^2)^3}=0.$$No conclusion about the limit.
If anyone could give me hints or point me in the right direction I would be grateful.
Thanks in advance.
 [/INST] Approaching along the path $x = y^2$ (motivated by trying to simplify $x^2 + y^4$, we have
$$\lim_{y \to 0} \frac{(y^2)^4 y^4}{((y^2)^2 + y^4)^3} = \lim_{y \to 0} \frac{y^{12}}{(2y^4)^3} = \frac 1 8 \ne 0.$$
Hence, the limit does not exist.

Alternative interpretation in polar coordinates: $\cos \theta \to 0$ in such a way that the denominator vanishes at the same order that the numerator does, and so the limit is non-zero.
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