Answer
The limit does not exist
Work Step by Step
Consider $f(x,y)=\dfrac{xy+1}{x^2-y^2}$
Let us consider the first approach: $(x,y) \to (1,-1)$ along $x=1$
Then, we get $\lim\limits_{y \to -1}\dfrac{y+1}{1-y^2}=\lim\limits_{y \to -1}\dfrac{(y+1)}{(1+y)(1-y)}=\dfrac{1}{1-(-1)}=\dfrac{1}{2}$
Let us consider the second approach: $(x,y) \to (1,-1)$ along $y=-1$
Then, we get $\lim\limits_{y \to -1}\dfrac{-x+1}{x^2-1}=\lim\limits_{y \to -1}\dfrac{(1-x)}{(x+1)(x-1)}=\dfrac{-1}{1+1}=-\dfrac{1}{2}$
This shows that there are different limit values when the approach is different and therefore, the limit does not exist for the given function.