jmc

Let $a=x+y$ and $b=1-xy.$ Then $$\begin{array}{rl}{a}^2+b^2& =(x+y{)}^2+(1-xy{)}^2\\ & =x^2+2xy+y^2+1-2xy+x^2{y}^2\\ & =1+x^2+y^2+x^2{y}^2\\ & =(1+x^2)(1+y^2),\end{array}$$so $$\frac{(x+y)(1-xy)}{(1+x^2)(1+y^2)}=\frac{ab}{a^2+b^2}.$$By AM-GM, $a^2+b^2\ge 2|ab|,$ so $$\left|\frac{(x+y)(1-xy)}{(1+x^2)(1+y^2)}\right|=\frac{|ab|}{a^2+b^2}\le \frac{1}{2}.$$Hence, $$-\frac{1}{2}\le \frac{(x+y)(1-xy)}{(1+x^2)(1+y^2)}\le \frac{1}{2}.$$Setting $y=0,$ the expression becomes $$\frac{x}{1+x^2}.$$As $x$ varies from $-1$ to 1, $\frac{x}{1+x^2}$ takes on every value from $-\frac{1}{2}$ to $\frac{1}{2}.$ Therefore, the set of all possible values of the given expression is $\boxed{\left[-\frac{1}{2},\frac{1}{2}\right]}.$