Question: Evaluate the following limit: $$\lim _{(x, y) \rightarrow(0,0)} \left(\frac{y^2}{x^4+y^2}=0.6\right)$$
I don't really know how to handle the fact that this is an implicit function and not the function $f(x,y) = \frac{y^2}{x^4+y^2}$. I'm guessing that you can show that the limit does not exist by showing that the limit along two different paths are different, but not quite sure.
Path 1: $x^2 = y$ \begin{align*} \lim _{\underset{\text{Along path } x^2 = y}{{(x, y) \rightarrow(0,0)}}} \frac{y^2}{x^4+y^2}=0.6 &\Leftrightarrow \lim _{y \rightarrow 0} \frac{y^2}{y^2+y^2}=0.6\\ &\Leftrightarrow \lim _{y \rightarrow 0} \frac{y^2}{2y^2}=0.6\\ &\Leftrightarrow \lim _{y \rightarrow 0} \frac{1}{2}=0.6\\ &\Leftrightarrow \frac{1}{2} = 0.6. \end{align*}
Path 2: $x = 0$ \begin{align*} \lim _{\underset{\text{Along path } x = 0}{{(x, y) \rightarrow(0,0)}}} \frac{y^2}{x^4+y^2}=0.6 &\Leftrightarrow \lim _{y \rightarrow 0} \frac{y^2}{0+y^2}=0.6\\ &\Leftrightarrow \lim _{y \rightarrow 0} \frac{y^2}{y^2}=0.6\\ &\Leftrightarrow \lim _{y \rightarrow 0} 1=0.6 \end{align*} It seems that something doesn't quite make sense as I end up with two numbers that are not $0.6$ equal to $0.6$. I was really confused how to proceed, but this is my best effort attempt.
You have found that the limit along two different paths is different, this means that the initial limit does not exist. Another simple path to choose is the path $y = 0$ which yields $0$ as a limit which again gives a different result.