Answer
Limit $l$ does not exist.
Work Step by Step
Consider $l=\lim\limits_{(x,y) \to (0,0)} \dfrac{y}{x^2-y}$
Suppose $y=p x^2$; $p \ne 1$
Now, $l=\lim\limits_{(x,y) \to (0,0)} \dfrac{y}{x^2-y}=\lim\limits_{(x,px^2) \to (0,0)} \dfrac{px^2}{x^2(1-p)}$
Thus, $l=\lim\limits_{(x,px^2) \to (0,0)} \dfrac{px^2}{x^2(1-p)}=\dfrac{p}{1-p}$
Hence, the limit $l$ does not exist as we can see that $l$ will have different limit for different values of $p$.
