Looks like the limit is
[tex]\displaystyle\lim_{(x,y)\to(0,0)}\frac{x^4-36y^2}{x^2+18y^2}[/tex]
Let [tex]y=x^m[/tex] with [tex]m\ge1[/tex], so the limit is equivalent to
[tex]\displaystyle\lim_{x\to0}\frac{x^4-36x^{2m}}{x^2+18x^{2m}}=\lim_{x\to0}\frac{x^2-36x^{2(m-1)}}{1+18x^{2(m-1)}}=0[/tex]
Now let [tex]x=y^n[/tex] with [tex]n\ge1[/tex], so the limit is
[tex]\displaystyle\lim_{y\to0}\frac{y^{4n}-36y^2}{y^{2n}+18y^2}=\lim_{y\to0}\frac{y^{2(2n-1)}-36}{y^{2(n-1)}+18}=-\dfrac{36}{18}=-2[/tex]
The value of the limit evidently depends on path of approach, so the limit does not exist.
