The function $F(x,y,z)$ on $(x,y,z)\in\mathbb{R}\times \mathbb{R}\times\left[0,\frac{1}{2}\right]$ is defined by \begin{equation} F(x,y,z)=\frac{x^2(1-2z)^2}{(1-z)^2}+2y^2-2(1-2z)^2 \end{equation} In general, if we want to find the implicit function $z(x,y)$ defined by $F(x,y,z)=0$, Implicit Function Theorem will fail to work because at some points (e.g., when $x=1,y=0$), there will be two solutions of $z$ ($z=0.5$ or $z=1-1/\sqrt{2}$) satisfying the equation $F(x,y,z)=0$.
Therefore, for any $(x,y)$, if $F(x,y,z)=0$ has no solution, then let $z(x,y)=0$; if it has only one solution, then we define $z(x,y)$ as that unique solution; if $F(x,y,z)=0$ has two solutions, we define $z(x,y)$ as the nontrivial one (the one other than $z=0.5$).
In this way, $z(x,y)$ is well-defined, but how to prove such function is continuous?
The above definition of the final function $z(x,y)$ assumes that every ordered pair $(x,y)$ with two images under the original function $z$ has the trivial image $\dfrac{1}{2}$, which is, then, ignored in the final definition. This is, however, not always true for the case $x = 0$ where the ordered pair $(0,y)$ gives the images:
$z = \dfrac{1-y}{2}$ and $z = \dfrac{1+y}{2}$
neither of which evaluate to $\dfrac{1}{2}$ for non-zero y, implying that the function is in fact not well-defined.
As for the problem in hand, consider the intersections of the surface with a plane of the type $x = a$, $y = a$ or $z = a$. Each such intersection is a curve and if the surface is continuous everywhere, all the curves obtained by substituting every possible value of a would also be continuous. Hope this helped.