Answer
$z$ is not a function of $x$ and $y$.
Work Step by Step
For $z$ to be a function of $x$ and $y$ then we require that for every ordered pair $(x,y)$ there is only one value of $z$ obtained from the formula connecting them.
Here we have
$$\frac{x^2}{4}+\frac{y^2}{9}+z^2=1\Rightarrow z^2=1-\frac{x^2}{4}+\frac{y^2}{9}$$
which yields
$$z=\pm\sqrt{1-\frac{x^2}{4}+\frac{y^2}{9}}.$$
Here we see that for some ordered pairs $(x,y)$ we will have two possible values for $z$, one with the plus and one with the minus sign so $z$ is NOT a function of $x$ and $y$.
