Let $x$ and $y$ be real numbers.  Find the set of possible values of
\[\frac{(x + y)(1 - xy)}{(1 + x^2)(1 + y^2)}.\]
Let $a = x + y$ and $b = 1 - xy.$  Then
\begin{align*}
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{align*}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 $\boxed{\left[ -\frac{1}{2}, \frac{1}{2} \right]}.$