System of equations: $$\begin{cases}\frac{x}{yz}+\frac{y}{zx}+\frac{z}{xy}=\frac{3}{xyz}\\xy+yz+zx=3\end{cases}$$ Solution:
$$\begin{cases}\frac{x}{yz}+\frac{y}{zx}+\frac{z}{xy}=\frac{3}{xyz}\\xy+yz+zx=3\end{cases}\Rightarrow\begin{cases}x^2+y^2+z^2=3\\xy+yz+zx=3\end{cases}$$
$x^{2}+y^{2}+z^{2}=xy+yz+zx$
$(x+y+z)^{2}-2(xy+yz+zx)=xy+yz+zx$
$(x+y+z)^{2}-2*3=3$
$(x+y+z)^{2}=9$
What should I do next?
Hint:
$$(x-y)^2+(y-z)^2+(z-x)^2=?$$
Alternatively, $$x^2-x(y+z)+y^2+z^2-yz=0$$
As $x$ is real, the discriminant must be $\ge0$
But $$(y+z)^2-4(y^2+z^2-yz)=-3(y-z)^2\le0$$
By the calculations you have made, your system is actually eqivalent to $$\begin{cases}x^2+y^2+z^2=3\\(x+y+z)^2=9\end{cases}$$Which is, at the same time, equivalent to the systems $$\begin{cases}x^2+y^2+z^2=3\\x+y+z=3\end{cases} \quad \begin{cases}x^2+y^2+z^2=3\\x+y+z=-3\end{cases}$$But you can check using the distance formula between a point and a plane (or in many other ways) that the planes $x+y+z=3$ and $x+y+z=-3$ are both tangent to the sphere $x^2+y^2+z^2=3$, so the only solutions are $x=y=z=\pm1$
For the system $\begin{cases}x^2+y^2+z^2=3\\x+y+z=3\end{cases}\quad$ just offset the variables $\begin{cases}x=1+X\\y=1+Y\\z=1+Z\end{cases}$
Then you get $\begin{cases}3+X+Y+Z=3\\3+2X+2Y+2Z+X^2+Y^2+Z^2=3\end{cases}\implies X^2+Y^2+Z^2=0$
Which in the real domain has only the $(0,0,0)$ solution.
For $x+y+z=-3$ proceed similarly with $x=X-1,y=Y-1,z=Z-1$.
