Range of $\frac{x-y}{1+x^2+y^2}=f(x,y)$ I have a function $\frac{x-y}{1+x^2+y^2}=f(x,y)$. And, I want to find the range of it. I analyzed this function by plotting it on a graph and found interesting things. Like if the level curve is $0=f(x,y)$, then I get $y=x$ which is a linear function. But if the level curve is something not 0, then the level curve becomes a circle. And for big values of level curves, the circle disappears. Is there something I can use to find the range of this function?

The denominator of the expression for the function $ \ z \ = \   f(x,y) \ = \  \frac{x-y}{1+x^2+y^2} \ $ has "four-fold" symmetry about the origin, which is "broken" by the "diagonal" (anti-)symmetry in the numerator.  So this function has the symmetry property $ \ f(x,-y) \ = \ -f(-x,y) \ \ $ and it is reasonable to expect that the extremal points of the function will be described by $ \ (\pm x \ , \ \mp x) \ \ . $
If we take "slices" $ \ x + y \ = \ c \ $ through the function surface, we obtain "cross-sectional" curves
$$  z \ \ = \ \  \frac{x \ - \ (c-x)}{1 \ + \ x^2 \ + \ (c-x)^2} \ \ = \ \ \frac{2x \ - \ c}{ 2x^2 \ - \ 2cx \ + \ (1 + c^2)} \ \ . $$
It is somewhat unappealing to differentiate this as it stands.  By completing-the-square in the denominator, we find that we can write
$$  z \ \ = \ \   \frac{2x \ - \ c}{ 2·\left(x^2 \ - \ cx \ + \ \frac{c^2}{4} \right) \ \ + \ (1 + c^2) \ - \ \frac{c^2}{2}} \ \ = \ \ \frac{2·\left(x \ - \ \frac{c}{2} \right)}{ 2·\left(x \ - \ \frac{c}{2} \right)^2 \ \ + \ \left(1 + \frac{c^2}{2} \right) } $$ $$ \rightarrow \ \ \frac{2·u}{ 2·u^2 \ \ + \ \left(1 + \frac{c^2}{2} \right) } \ \ . $$
It is clear that the denominator of the expression is (still) never equal to zero, so we can establish that the derivative of this transformed function is zero for $ \ 4·u^2 \  + \ 2·\left(1 + \frac{c^2}{2} \right) \ - \ 2u·4u \ = \ 0 $ $ \Rightarrow \ u^2 \ = \ \frac12·\left(1 + \frac{c^2}{2} \right) \ \ . $  This locates the extremal points on the cross-sectional curve, with the extremal values being
$$ z \ \ = \ \frac{\pm \ \sqrt2 \ · \ \sqrt{1 + \frac{c^2}{2}}}{ \left(1 + \frac{c^2}{2} \right) \  + \ \left(1 + \frac{c^2}{2} \right) } \ \ = \ \ \pm \frac{  \sqrt2}{2} \ · \ \frac{1}{\sqrt{1 + \frac{c^2}{2}}} \ \ = \ \  \frac{\pm \ 1}{\sqrt{2 \ + \ c^2 }} \ \ . $$
We see from this result that $ \ \lim_{|c| \ \rightarrow \ \pm \infty} \ \ z \ = \ 0 \ \ , $ as you observed.  The extremal values are greatest in absolute-value for $ \ c = 0 \ \ , $ confirming that the extremal points occur on the line $ \ x + y \ = \ 0 \ \ . $ The absolute maxima and minima for the function are thus $ \ \pm \frac{ 1}{\sqrt2} \ \ . $
[We did not need to know the locations of the extrema, but we use our relations at $ \ c = 0 \ $ to find $ \ u \ = \ x - \frac{0}{2} \ = \ x \ = \ \pm \frac{ 1}{\sqrt2} \ $ and $ \ y = \ 0 - x \ = \ \mp \frac{ 1}{\sqrt2} \ \ . \ ] $
