System of non-linear equations (ISNMO, 1991) Let $x,y,z$ be real numbers. Solve the following system of equations: $$\begin{cases}
\frac{3(x^2+1)}x=\frac{4(y^2+1)}y=\frac{5(z^2+1)}z;   \\
xy+xz+yz=1.
\end{cases}$$
I tried to solve this system using the following method:
Since $$\frac{3(x^2+1)}x=\frac{4(y^2+1)}y; \Rightarrow 3y(x^2+1)=4x(y^2+1).$$ Now, from the second row, we have $$3y(x^2+xy+xz+yz)=4x(y^2+xy+xz+yz); \Rightarrow (x+y)(xy+4xz-3yz)=0.$$
Thus, $$x=-y; z=\frac {xy}{3y-4x}.$$
But if we replace $y$ with $-x$, we obtain a complex solution. The same happens when we replace $z$ with $\frac {xy}{3y-4x}$ in the second row because it takes us back to the first expression $x=-y$.
Finally, in case we try to deal with $$\frac{3(x^2+1)}x=\frac{5(z^2+1)}z, z=\frac {xy}{3y-4x},$$
then it complicates the problem with huge expressions.
Wolfram Alpha shows that the real solutions are $(x;y;z)=(\frac 13;\frac 12;1)\cup(-\frac 13;-\frac 12;-1)$, but I do not know how to get to them. I would appreciate any help.
Thank you in advance.

HINT.-Equalizing to $t$ each of the expressions in first line we have
$$3(x^2+1)=tx\\4(y^2+1)=ty\\5(z^2+1)=tz$$ we deduce
$$x=\frac{t\pm\sqrt{t^2-36}}{6}\\y=\frac{t\pm\sqrt{t^2-64}}{8}\\z=\frac{t\pm\sqrt{t^2-100}}{10}$$Now taking into account the second line of the proposed system we have the solution given by Wolfram:
$$\left(\frac{t\pm\sqrt{t^2-36}}{6}\right)\left(\frac{t\pm\sqrt{t^2-64}}{8}\right)+\left(\frac{t\pm\sqrt{t^2-36}}{6}+\frac{t\pm\sqrt{t^2-64}}{8}\right)\left(\frac{t\pm\sqrt{t^2-100}}{10}\right)=1$$ but it is not necessary if we before that $t=10$ is a good value.
