Answer
Please see the figure attached.
Notes:
- Red dashed curves are the vertical traces parallel to the $xz$-plane
- Black dashed curves are the horizontal traces
Work Step by Step
We have $f\left( {x,y} \right) = \sqrt {4 - {x^2} - {y^2}} $. This is the equation of the upper hemisphere of radius $2$. By setting $y=a$ we fix the $y$-coordinate and obtain the vertical trace curve $f\left( {x,a} \right) = z = \sqrt {4 - {x^2} - {a^2}} $ that lies in the plane parallel to the $xz$-plane. Similarly, we can set $x=a$ to fix the $x$-coordinate and obtain the vertical trace curve $f\left( {a,y} \right) = z = \sqrt {4 - {a^2} - {y^2}} $ that lies in the plane parallel to the $yz$-plane.
The horizontal traces are the curves $c = \sqrt {4 - {x^2} - {y^2}} $. So,
${c^2} = 4 - {x^2} - {y^2}$,
${y^2} = 4 - {x^2} - {c^2}$,
$y = \pm \sqrt {4 - {x^2} - {c^2}} $
