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 $f\left( {x,y} \right) = {x^2} + 4{y^2}$. By setting $y=a$ we fix the $y$-coordinate and obtain the vertical trace curve $f\left( {x,a} \right) = z = {x^2} + 4{a^2}$ 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 $f\left( {a,y} \right) = z = {a^2} + 4{y^2}$ that lies in the plane parallel to the $yz$-plane.
The horizontal traces can be obtained by setting $c = {x^2} + 4{y^2}$. These are equations of the ellipses in standard position.
$c = {x^2} + 4{y^2}$,
$4{y^2} = c - {x^2}$,
$y = \pm \frac{1}{2}\sqrt {c - {x^2}} $
