Answer
$z$ is not a function of $x$ and $y$.
Work Step by Step
For $z$ to be the function of $x$ and $y$ we need that for every ordered pair $(x,y)$ there be only one value for $z$.
Here we have
$$xz^2+2xy-y^2=4$$
Solving for $z$ we have
$$xz^2+2xy-y^2=4\Rightarrow xz^2=4+y^2-2xy\Rightarrow z^2=\frac{4+y^2-2xy}{x}$$
and this finally gives us
$$z=\pm\sqrt{\frac{4+y^2-2xy}{x}}.$$
This says that there are ordered pairs $(x,y)$ for which $z$ will have TWO VALUES-one obtained putting plus and the other putting minus. Thus $z$ is NOT a function of $x$ and $y$.
