Starting from any $(x_0,y_0,z_0)\in \mathbb{C}^3$, can the following ODE system have a solution for all real number?
\begin{align} x'(t) &=3 y^2(t) \\ y'(t) &=2 x(t) z(t)-1 \\ z'(t) &=0 \end{align} $$ x(0)=x_0, \quad y(0)=y_0, \quad z(0)=z_0 $$ I think it can not be solved for all $\mathbb{R}$ for some starting points. How to show an ODE system has no global solution?
On
This is not a full answer, but is larger than a comment so I have decided to post it as an answer.
If $z'=0$ then $z(t) = z(0) = z_0$ (globally).
If $z_0 = 0$ then $y'(t) = -1$, so $y(t) = -t + y_0$ (globally). Next, $x'(t) = 3 (-t + y_0)^2$ and this quickly gives you the global solution for $x$. So, in this case your system has a unique global solution.
Assume now that $z_0 \ne 0$.
Let $X(t) = \int \limits _0 ^t x(s) \Bbb d s$, for $t$ in some small neighbourhood of $0 \in \Bbb R$. Integrating with respect to $t$ in the equation $y'(s) = 2 x(s) z_0 -1$ you get $y(t) - y_0 = 2 z_0 X(t) -t$. Take this formula for $y(t)$ and plug it into the first equation, obtaining $X''(t) = x'(t) = 3 (y_0 + 2 z_0 X(t) -t)^2$.
Let $U(t) = y_0 + 2z_0 X(t) - t$ and note that $U(0) = y_0 + 2 z_0 X(0) = y_0$, that $U'(0) = 2 z_0 x_0 -1$ and that $U''(t) = 2 z_0 X''(t)$. Then the first equation can be rewritten as $\frac 1 {2 z_0} U''(t) = 3 U(t)^2$, or $U''(t) = 6 z_0 U(t)^2$.
Here things become shaky.
Observe that $\big( (U')^2 \big)' = 2 U' U'' = 2 U' 6 z_0 U^2 = 4 z_0 (U^3)'$, so $(U')^2 = 4 z_0 U^3 + C_1$ with $C_1$ some integration constant. Make the change of variable $s = t \sqrt {z_0}$ and let $U(t) = V(s) = V(t \sqrt {z_0})$, so $U'(t) = V'(s) \sqrt {z_0}$ and your equation becomes $(V')^2 = 4 V^3 + D_1$, with $D_1 = \frac {C_1} {z_0}$. Remarkably, this equation is well known and its solution is $V(s) = \wp (s; \omega_1, \omega_2)$ corresponding to the elliptic invariants $g_2 = 0$ and $g_3 = -D_1$. It follows that $U(t) = \wp (t \sqrt {z_0}, \omega_1, \omega_2)$, so $X(t) = \big( \wp (t \sqrt {z_0}, \omega_1, \omega_2) + t -y_0 \big) \frac 1 {2 z_0}$ and then $x(t) = \big( \wp ' (\sqrt {z_0} t, \omega_1, \omega_2) \sqrt {z_0} +1 \big) \frac 1 {2 z_0}$.
Unfortunately, I have no experience in working with elliptic functions. Furthermore, I am not sure about the change of variables from $t$ to $s$ since this takes us from derivable functions of a real variable to holomorphic functions of a complex variable. I also do not know how to deduce the periods from the elliptic invariants, and what is the relation between them and the initial conditions for $U$. Hopefully, someone more experienced than me will take it over from here, either by editing and completing my answer, or by adding another answer.
Also, I find strange that Mathematica gives a different solution for $U$, namely $\frac 1 {\sqrt[3] {z_0} } \wp (\sqrt[3] {z_0} (t+C_1); 0, C_2)$, with $C_1, C_2$ two integration constants (not related to my $C_1$ from above). When also adding the initial conditions to the mix, Mathematica complains about them, but I do not understand its error message: "For some branches of the general solution, unable to solve the conditions." It seems to confirm the OP's suspicions.
If $c=0$, then $x=(t-a)^3+b$, $y=a-t$ is the general (global) solution. If $c\ne0$ then $x=1/(2\,c)$, $y=0$ is a global solution. So the system has global solutions.