After setting all three equations to $0$, multiply the first equation by $z$, the second by $x$, and the third by $xy$ to get:

\begin{align}
2xz + 2xyz &= 0 \quad (1) \\
x^3 + 2xyz &= 0 \quad (2) \\
xy^3 + 2xyz - 4xy &= 0 \quad (3) \\
\end{align}

\begin{align}
(1) - (2) &\Rightarrow 2xz - x^3 = 0 &(4) \\
(2) - (3) &\Rightarrow x^3 - xy^3 -4xy = 0 &(5) \\
(1) - (3) &\Rightarrow 2xz - xy^3 -4xy = 0 &(6) \\
\end{align}

Subtracting $(6)$ from $(5)$ gives $x^3 - 2xz = x(x^2-2z) = 0$ so that $x = 0$ or $z = x^2/2$.

In the first case, we get $2yz=0$ from the second equation, so either $y=0$ or $z=0$.

If $y=0$, then $2z-4=0$ so that $z=2$, which gives the solution $(0,0,2)$.

If $z=0$ then $y^2 = 4$ so that $y=\pm2$, which gives the solutions $(0,2,0)$, and $(0,-2,0)$.

In the second case, we get $x\neq0$ so that $z=x^2$.

Then, first original equation gives $2x(1+y) = 0$. But $x \neq 0$, so we must have $y=-1$.

Plug this into the third original equation to get $(-1)^2 + 2z - 4 = 0$, i.e. $z = \frac{3}{2}$.

Now plug $y=-1$ and $z=3/2$ into the second original equation to get $x^2 +2(-1)(\frac{3}{2}) = 0$ which gives $x^2 = 3$ or $x = \pm\sqrt 3$.

This gives the final solutions, $(-\sqrt 3, -1, \frac{3}{2})$, and $(\sqrt 3, -1, \frac{3}{2})$.

So there are three solutions: $(0, 0, 2)$, $(0,-2,0)$, $(0, 2, 0)$, $(-\sqrt 3, -1, \frac{3}{2})$, and $(\sqrt 3, -1, \frac{3}{2})$. You can verify that these work by plugging them in to the original equations.

## Best Answer

HintLet $p=u^3-v^4-4,\ q=2u^2-3v^4-8,$ so that your system is $$x^2-y^2=p, \\ 2xy+y^2=q. \tag{1}$$ If the second equation of $(1)$ is solved for $x$ one has $$x=\frac{q-y^2}{2y}. \tag{2}$$ Now substitute this into the first equation of $(1)$ and move $p$ to the left side, obtaining an expression which must be zero. This expression has a denominator of $4y^2$ which is irrelevant for solutions, and the negative of its numerator is quadratic in $y^2$, i.e. only contains $y^4,y^2$ [and $p,q$ ] so it determines $y^2$ from $p,q.$ Specifically, $$3y^4+2(2p+q)y^2-q^2=0. \tag{3}$$ Now here's why this answer is only a hint: at this point you may be able to use the initial conditions that at $(u,v)=(2,1)$ you have $(x,y)=(2,-1).$ I did calculate that at these values of $u,v$ one has $p=3,\ q=-3$ and when plugged in this agrees with $(x,y)=(2,-1).$ When $(3)$ is solved for $y^2$ there will be a sign choice on the $y^2$, and after that on passing to $y$ itself there is another sign choice, for a total of four sign choices in all. To make the solution local satisfying the conditions, the right sign choices need to be made, which determines $y,$ and then $x$ is forced by $(2).$Given the forms of $p,q$ this approach may give rather complicated formulas for $x,y$ in terms of $u,v$. There may be a simpler way but I don't see one.