Answer
A parametrization of the intersection is
${\bf{r}}\left( t \right) = \left( {2\cosh t,2\sinh t,2\sinh 2t} \right)$
Work Step by Step
Suppose there exists some parameter $t$ such that $x = 2\cosh t$ and $y = 2\sinh t$.
Since ${\cosh ^2}t - {\sinh ^2}t = 1$, so
${x^2} - {y^2} = 4{\cosh ^2}t - 4{\sinh ^2}t = 4$
Thus, the $x$- and $y$-coordinates of points on the surface ${x^2} - {y^2} = 4$ can be represented by the parametrization $\left( {x,y} \right) = \left( {2\cosh t,2\sinh t} \right)$.
Hence, the intersection of the surfaces ${x^2} - {y^2} = 4$ and $z = xy$ can be parametrized by
${\bf{r}}\left( t \right) = \left( {2\cosh t,2\sinh t,4\cosh t\sinh t} \right)$
Since $\sinh 2t = 2\sinh t\cosh t$, so
${\bf{r}}\left( t \right) = \left( {2\cosh t,2\sinh t,2\sinh 2t} \right)$
