1.st case: If $x,y,z \ge 0$, then from Am-Gm we have: $(2x^3+1)(2y^3+1)(2z^3+1) \ge 27x^2y^2z^2$, so the only solution is: $x=y=z=1$. 2.nd case: If exactly one of $x,y,z$ is greater than 0 (WLOG we can assume that is $x$), then we have: $2x^3+1>0>3xz$ what is contradiction, so we don't have solutions in this case. 3.rd case: If exactly two of $x,y,z$ are greater than (WLOG we can assume they are $x$ and $y$), then we have: $2x^3+1>0>3xz$ what is contradiction, so we don't have solution in this case ether. 4.th case: If $x,y,z<0$, we can suppose that $x=\max\{x,y,z\}$, then we have: $1+2x^3 \ge 1+2z^3 \Rightarrow 3xz\ge 3yz \Leftrightarrow x\le y$ what can be true with assumption just when $x=y$. That means that $x=y\ge z$, what implies: $2y^3+1 \ge 2z^3+1 \Rightarrow 3xy \ge 3yz \Leftrightarrow x\leq z$ what can be true with assumption just if $x=z$. Hence, in this case, system of equations can have solution just when $x=y=z$, but then it becomes: $2x^3+1=3x^2$ what is equivalent to: $(x-1)^2(2x+1)=0$. That means that only solution in this case is $x=y=z= \frac{-1}{2}$.