Austria competition problems, 2010

We first note that
$$\begin{aligned} & 4x^4 - x^2(4y^4 + 4z^4 - 1) - 2xyz + y^8 + 2y^4z^4 + y^2z^2 + z^8 \\ & = (4x^4 + y^8 + z^8 - 4x^2y^4 - 4x^2z^4 + 2y^4z^4) + (x^2 - 2xyz + y^2z^2) \\ & = (2x^2 - y^4 - z^4)^2 + (x - yz)^2. \end{aligned}$$
The given equation is therefore equivalent to
$$(2x^2 - y^4 - z^4)^2 + (x - yz)^2 = 0.$$
It therefore follows that both $x = yz$ and $2x^2 - y^4 - z^4 = 0$ must hold.
Substituting $x = yz$ in the second of these equations, we obtain $2y^2z^2 - y^4 - z^4 = 0$, which is equivalent to $-(y^2 - z^2)^2 = 0$. We see that $z = \pm y$ must hold. For $z = y = t$, we obtain $x = t^2$, and for $-z = y = t$, we obtain $x = -t^2$. It therefore follows that the set of all solutions is
$$\{ (t^2, t, t) \mid t \in \mathbb{R} \} \cup \{ (-t^2, t, -t) \mid t \in \mathbb{R} \}.$$