Is this the same "f(x,y)" as in (a)? If so, doesn't (a) already ask you to show that this limit doesnotexist? Are you sure you have copied these problems correctly? Perhaps this was the limit as (x, y) goes to (0, -1)?

I recommend you start by doing what you were told to do, " Using the paths x=0 and y=x+1"!c) Is it possible to define a new function g(x,y) that is defined and continuous for all (x,y) in R^2, and such that (9x,y)=f(x,y) for all (x,y) in the domain of f? If so, find such a function, If not, explain why not.

I'm not sure where to start for a)

On x=0, , as long as y is not 1 or -1, for all x so the limit, as y goes to 1, is 0.

On y= x+ 1, x= y- 1 so and

What?? You've set x= 1 and y= 0! (0, 1) means x= 0 and y= 1.for b) i think its

I assume the "9" is a typo. Do you mean g(x,y)= f(x,y)? Since f(x,y) is defined for all (x,y) except (x, 1) and (x, -1), such a function would have to have the same limit as (x, y) goes to (x, 1) or as (x, y) goes to (x, -1), for any x. And that is possible only if that limit exist.and I also don't know where to start for c)