I've been trying to paramaterize
$$x^2+y^2+z^2=9,\ x^2-y^2=3$$
but haven't had any luck. I was thinking to let $x=\sqrt{3}\ \sec(t), y=\sqrt{3}\ \tan(t)$ to complete the identity on the right equation, but when going back to the left one and solving for $z$, I get a nonreal result.
The sphere equation becomes
$$ z^2 \ = \ 9 \ - \ 3 \sec^2 t \ - \ 3 \tan^2 t $$
$$ = \ 9 \ - \ 3 \ (\tan^2 t \ + \ 1 ) - \ 3 \tan^2 t \ = \ 6 \ - \ 6 \tan^2 t \ . $$
Doesn't seem like there's anything problematic there, since $ \ |z| \ \le \ 3 \ $ ; the upper bound on $ \ z^2 \ $ of 6 represents the greatest "altitude" from the $ \ xy-$ plane where the surfaces intersect. There will be two square-roots, corresponding to locations on the two hemispheres about the $ \ xy-$ plane. (I'm adding a picture of the surfaces.)