Systems of equations involving linear and quadratic terms

Multiply each equation by $2$ and add all of them to get $$(x-y)^2+(y-z)^2+(z-x)^2=24.$$ Let $u=x-y, v=y-z, w=z-x$. Then you have \begin{align*} u+v+w&=0\\ u^2+v^2+w^2&=24 \end{align*} Now in your original system of equations, subtract equ 2 from equ 1, equ 3 from equ 2 and equ 1 from equ 3 to get \begin{align*} (x-y)(x+y+z) &=-1\\ (y-z)(x+y+z) &=-1\\ (z-x)(x+y+z) &=2 \end{align*} Clearly $x+y+z \neq 0$, so from here we can conclude that \begin{align*} x-y&=y-z \implies u=v\\ z-x&=-2(y-z) \implies w=-2v\\ \end{align*}

So now plug this in the $u,v,w$ equation s above to get $$(v)^2+(v)^2+(-2v)^2=24 \implies v =\pm 2.$$ Hopefully now you can solve the rest.

We have $$\left\{ \begin{aligned} x^2y - y^2z &= 3y \\ y^2z - xz^2 &= 4z \\ z^2x - x^2y &= 5x \end{aligned} \right. $$ and after summing we obtain: $$3y+4z+5x=0.$$ Also, $$ \left\{ \begin{aligned} x^2z - yz^2 &= 3z \\ y^2x - x^2z &= 4x \\ z^2y - xy^2 &= 5y \end{aligned} \right. $$ and after summing again we obtain: $$3z+4x+5y=0.$$ The rest is smooth:

From $$5x+3y+4z=0$$ and $$4x+5y+3z=0$$ we obtain: $$y=-\frac{x}{11}$$ and $$z=-\frac{13x}{11},$$ which gives $$x^2-\frac{13x^2}{121}=3$$ and from here $$x=\pm\frac{11}{6}.$$

I got the following answer: $$\left\{\left(\frac{11}{6},-\frac{1}{6},-\frac{13}{6}\right),\left(-\frac{11}{6},\frac{1}{6},\frac{13}{6}\right)\right\}.$$