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?

$\frac{x-y}{1+x^2+y^2}-z = 0$
You can find $\frac{\partial f}{\partial x} = 0$ and $\frac{\partial f}{\partial y} = 0$
Or you can find the direction of normal to the surface(z) and the normal must have direction ratios as $\langle0\hat{i}+0\hat{j}±1\hat{k} \rangle$
$\nabla f(x,y,z)=$$\frac{2x(x-y)-(1+x^2+y^2)}{(1+x^2+y^2)^2}\hat{i}+\frac{2y(x-y)+(1+x^2+y^2)}{(1+x^2+y^2)^2}\hat{j}-\hat{k}$
Let us put $y = -x$
$2x(2x)-(1+x^2+x^2) = 0$
Solving above:
{$x,y$}  ≡ {($2^{-1/2}, -2^{-1/2}$), $(-2^{-1/2},2^{-1/2} $)}
$f(2^{-1/2}, -2^{-1/2})$, $f(-2^{-1/2},2^{-1/2})$
implies the function has only two such maxima and minima $\frac{1}{\sqrt{2}}$ and $\frac{-1}{\sqrt{2}}$
As, the denominator dominates over the numerators (Quadratic in nature)
The range of function: [$\frac{-1}{\sqrt{2}}$, $\frac{1}{\sqrt{2}}$]
