Does $lim_{(x,y)to (a,b)} frac{f(x)}{g(x,y)}$ that equals to $lim_{(x,y)to (a,b)} frac{g(x,y)}{f(x)}$ mean that the limit equals to 1?

Firstly observe that by $x-x_i=u$ and $y-_i=v$ we have

$$lim_{x to x_i, y to y_i} frac{x-x_i}{sqrt{(x-x_i)^2 + (y-y_i)^2}} =lim_{(u,v)to (0,0)} frac{u}{sqrt{u^2 + v^2}} $$

which doesn't exist (just consider path $u=v=tto 0^+$ and path $u=v=tto 0^-$).

For the general question when

$$lim_{(x,y)to (a,b)} F(x,y)=Lneq 0$$

exists then by reciprocal law for limits

$$lim_{(x,y)to (a,b)} frac{1}{F(x,y)}=frac 1Lneq 0$$

and thus when both limits exist

$$lim_{(x,y)to (a,b)} F(x,y)=lim_{(x,y)to (a,b)} frac{1}{F(x,y)}$$

implies $L^2=1$.

Hint: when we have $a = frac{1}{a}$, then it gives $a^2 = 1$.

L'hopital rule is for one variable not multiple variables, you have a problem lim (x->x0) lim(y->y0) f(x,y) =/= lim(y->y0) lim (x->x0) f(x,y)

in the first case the limit is 1 and in the second case lim(y->y0) sqrt((y-y0)^2)/0 which is infinite.

So the answer is no, the limit for multiple variables you need to transform those variables into the polar coordinates and proof that a limit doesn't depend on particular a vector direction.