Show that $\lim_{(x, y)\to(0,0)}\frac{x^4y^4}{(x^2+y^4)^3}$ D.N.E Show that $$\lim_{(x,y)\to(0,0)}\frac{x^4y^4}{(x^2+y^4)^3}$$ D.N.E
In order to do this I should consider two paths and show that they do not reach the same limit.
Consider when $x=0$. 
Then we have $\lim_{y\to 0}\dfrac{y^4}{y^{12}}=\dfrac{1}{y^8}=\infty$ 
Similarly if we choose $y=0$, then we have
Then we have $\lim_{x\to 0}\dfrac{x^4}{x^{6}}=\dfrac{1}{x^2}=\infty$ 
Since both of these limits approach an asympote, can I even use them to show that the limit does not exist? Or must they approach a finite number?

Write the function as $$\frac{x^4 y^4}{(x^2 + y^4)^3} = \frac{x^4 y^4 / x^6}{(1 + (y^4/x^2))^3} = \frac{y^4/x^2}{(1+(y^4/x^2))^3}.$$  Now we can see that if $(x,y) \to (0,0)$ in such a way that the ratio $y^4/x^2$ does not tend to $0$, the resulting limit will not be zero.  This furnishes the motivation behind the choice of path $x = y^2$.  In fact, we can see that for any real constant $c$, if we choose the parametrization $(x,y) = (t^2, ct)$, then $y^4/x^2 = c^4$ is independent of the parameter $t$, and $$\lim_{(x,y) \to (0,0)} \frac{x^4 y^4}{(x^2 + y^4)^3} = \lim_{t \to 0} \frac{c^4}{(1+c^4)^3} = \frac{c^4}{(1+c^4)^3}.$$
