We should find discriminants from the first equation. Let $D_{x}, D_{y}, D_{z}$ be discriminants for $x, y, z$ respectively. Now we will find them: $D_{x}= b^2-4ac=-4y(y^2z+z^2)=-4y^3z-4yz^2$ $D_{y}= b^2-4ac=x^4-4z\cdot z^2=x^4-4z^3$ $D_{z}= b^2-4ac=y^4-4x^2y$ Now, $D_{x} \geq 0, D_{y} \geq 0, D_{z} \geq 0$, so we have a system of inequalities: $-4y^3z-4yz^2 \geq 0$ $x^4-4z^3 \geq 0$ $y^4-4x^2y \geq 0$ And when we add them we get $\frac{1}{4}(x^4+y^4) \geq z^3 + z^2 y + z y^3 + x^2 y$ which is same as our second condition, so we now have a system of equations from which we get that: $z=-y^2, x^4=4z^3, y^3=4x^2$. Now it is easy to get $\frac{y^6}{16}=-4y^6$ which holds only for $y=0$. Finally $x=y=z=0$ is the only solution.

Notice we can derive two relations from the first equation, which are: $(I): (z + \frac{y^2}{2})^2 = \frac{y^4}{4} - x^2y$ $(II): (\frac{x^2}{2} + yz)^2 = \frac{x^4}{4} - z^3$ Summing these two, we get: $(z + \frac{y^2}{2})^2 + (\frac{x^2}{2} + yz)^2 = \frac{x^4 + y^4}{4} - z^3 -x^2y = z^2y + zy^3 = -(xy)^2$ Hence, $(z + \frac{y^2}{2})^2 + (\frac{x^2}{2} + yz)^2 + (xy)^2 = 0$ and now is just conclude that $x=y=z=0$

Even though it's very contrived, I find this problem hilarious. Look at the first equation, where we have \begin{align*} \Delta_x \geq 0 &\implies y^3z+yz^2 \leq 0 \\ \Delta_y \geq 0 &\implies x^4-4z^3 \geq 0 \\ \Delta_z \geq 0 &\implies y^4-4x^2 y \geq 0. \end{align*}Summing $\frac 14$ times the second and third equations and subtracting the first, we yield the second equality as an equality. Thus, equality must hold in all three of these inequalities, so $x=y=z=0$.

We can look at the discriminants when the first equation is considered as a quadratic in $x$, $y$, or $z$, which gives us the inequalities \begin{align*}z^2y+zy^3&\le 0 \\ x^4-4z^3&\ge 0 \\ y^4-4x^2y&\ge 0. \end{align*}Now look at these, considering the second equation. Note that $\frac{1}{4}(x^4+y^4)\ge x^2y+z^3$. So we can cancel out those terms, and the remaining inequality is \[z^2y+zy^3\ge 0.\]However, from our above discriminant-derived inequalites, $z^2y+zy^3\le 0$, which means $yz(y^2+z)=0$. Assume now that none of $x$, $y$, or $z$ is $0$. So $z=-y^2$, so $x^4\ge -4y^6$ and $x^2\le \frac{y^3}{4}$. Then $\frac{y^6}{16}\le -4y^6$, clearly impossible unless $y=0$. So the only solution is $(x, y, z)=(0, 0, 0)$. $\blacksquare$

We claim that the only solution is $(x, y, z) = (0, 0, 0)$. This clearly works. Consider the discriminant of the first equation as a quadratic in $x, y, z$ respectively. We get $$\Delta_x=-4y(zy^2+z^2) \ge 0 \Rightarrow 0 \ge zy^3+z^2y$$$$\Delta_y=x^4-4z^3 \ge 0 \Rightarrow x^4 \ge 4z^3$$$$\Delta_z=y^4-4x^2y \ge 0 \Rightarrow y^4 \ge 4x^2y.$$Adding the last two inequalities, $$\frac14(x^4+y^4) \ge z^3+x^2y.$$But by the second equation, this means that the equality case of the first inequality must hold. From this we deduce that $z=-y^2$, but plugging this into the first original equation yields $x^2y=0$, so either $x=0$ or $y=0$. We can check that both cases do indeed give $(x, y, z)=(0, 0, 0)$, as desired.

Viewing the first equation as a quadratic in $x,y,z$, we get \begin{align*} \Delta_x &= -4(y^3z+yz^2) \ge 0 \\ \Delta y &= x^4-4z^3 \ge 0 \\ \Delta z &= y^4-4x^2y \ge 0 \end{align*} Adding the equations and rearranging we get the second equality as an inequality, implying that $x=y=z=0$. Thus, the only ordered pair is $\boxed{(0,0,0)}$.

what in the world The only solution is $(x, y, z) = (0, 0, 0)$. This clearly works. View the first equation as a quadratic in $x$, and let $D_x$ be the discriminant; then $D_x = -4y^3z - 4yz^2$. Defining $D_y$ and $D_z$ similarly, we have $D_y = x^4 - 4z^3, D_z = y^4 - 4x^2y$. For $x$, $y$, and $z$ to be real, we must have $D_x, D_y, D_z \ge 0$; but the second equation rewrites as $$x^4 + y^4 - 4z^3 - 4z^2y - 4zy^3 - 4x^2y = 0 \implies D_x + D_y + D_z = 0,$$so in fact we have equality and $D_x = D_y = D_z = 0$. Looking at the equation for $D_x$, we have $yz(y^2 - z) = 0$. Then If $y = 0$, the original first equation gives $z = 0$, and then the equation for $D_y$ gives $x = 0$ If $z = 0$, the equation for $D_y$ gives $x = 0$, and the original first equation gives $y = 0$ If $y^2 = z$, then the equation for $D_y$ gives $x^2 = \pm 2y^3$, so from the third equation we have $y = 0$, which we already handled Thus $(0, 0, 0)$ is indeed the only solution.

Consider the first equation as a quadratic wrt $x$, $y$, and $z$. Let $\Delta_{\chi}$ be the discriminant of a quadratic in variable $\chi$. Then \begin{align*} \\ \Delta_x = -4y^3z - 4yz^2 \geq 0 \implies 0 \geq zy(y^2 + z) \\ \Delta_y = x^4 - 4z^3 \geq 0 \implies x^4 \geq 4z^3 \\ \Delta_z = y^4 - 4x^2y \geq 0 \implies y^4 \geq 4x^2y \\ \end{align*} Summing the last two inequalities gives us \[x^4 + y^4 \geq 4(z^3 + x^2y) \implies \frac{1}{4}(x^4 + y^4) \geq z^3 + x^2y\]And adding the last inequality gives \[\frac{1}{4}(x^4 + y^4) \geq z^3 + x^2y + z^2y + zy^3\]And since this is the second original equality, we get that $x = y = z = 0$ must be the only solution.