For $g(X) = \frac{(x^2+y^4)^3}{1+x^6y^4}$ , show $\lim\limits_{|x| \rightarrow \infty} g(x,ax) = \infty? $ for any real number $a$ 
Let $$g(X) = \frac{(x^2+y^4)^3}{1+x^6y^4}.$$ (1) Show that $\lim\limits_{|x| \rightarrow \infty} g(x,ax) =\infty $ for any real number $a$.
(2) Does $\lim\limits_{|X| \rightarrow \infty } g(X) = \infty$?

(1) Considering $g(x,ax)$:
$$g(x,ax) = \frac{(x^2+(ax)^4)^3}{1+x^6(ax)^4}= \frac {x^6+3a^4x^8+3x^{10}a^8+a^{12}x^{12}} {1+x^{10}a^4} = \frac {\frac{1}{x^4}+\frac{3a^4}{x^2}+3a^8+a^{12}x^{2}} {\frac{1}{x^{10}}+a^4}$$
It follows that as $|x| \rightarrow \infty$, we have $g(x,ax) \rightarrow \infty$
(2) Does $\lim\limits_{|X| \rightarrow \infty } g(X) = \infty$?
$$g(X) = \frac{(x^2+y^4)^3}{1+x^6y^4}.$$ I believe I am supposed to factor out $|X|^2=|x^2+y^2|$ but I do see how.
I d like to know if I am in the right direction. What would be a $\epsilon,\delta$ approach is this case?
Much appreciated for your help/explanation.

For (2),
let
$x=t^u, y=t^v$
and I will try to choose
$u$ and $v$
to make things interesting.
$\begin{array}\\
g(X) 
&= \dfrac{(x^2+y^4)^3}{1+x^6y^4}\\
&= \dfrac{(t^{2u}+t^{4v})^3}{1+t^{6u}t^{4v}}\\
&\sim \dfrac{(t^{\max(2u, 4v)})^3}{t^{6u+4v}}\\
&= \dfrac{t^{\max(6u, 12v)}}{t^{6u+4v}}\\
&= t^{\max(6u, 12v)-6u-4v}\\
\end{array}
$
If $12 v < 6u+4v$,
then $g(X) \to 0$.
This is
$8v < 6u$.
This works if,
for example,
$u=2, v=1$.
Try
$x = t^2, y=t$.
Then
$g(X)
=\dfrac{(t^4+t^4)^3}{1+t^{12}t^4}
=\dfrac{8t^{12}}{1+t^{16}}
\to 0
$.
