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 $f\left( {x,y} \right) = \frac{1}{{{x^2} + {y^2} + 1}}$. By setting $y=a$ we fix the $y$-coordinate and obtain the vertical trace (red dashed curve in the figure) $f\left( {x,a} \right) = z = \frac{1}{{{x^2} + {a^2} + 1}}$ that lies in the plane parallel to the $xz$-plane. Similarly, if we set $x=a$, we obtain the vertical trace $f\left( {a,y} \right) = z = \frac{1}{{{a^2} + {y^2} + 1}}$ that lies in the plane parallel to the $yz$-plane.
The horizontal traces can be obtained by setting $c = \frac{1}{{{x^2} + {y^2} + 1}}$ (black dashed curves in the figure).
${x^2} + {y^2} + 1 = \frac{1}{c}$
$y = \pm \sqrt {\frac{1}{c} - 1 - {x^2}} $
