378*. Solve the system of equations:

$$
\left\{\begin{array}{l}
x+y=\frac{5 x y}{1+x y} \\
y+z=\frac{6 y z}{1+y z} \\
z+x=\frac{7 z x}{1+z x}
\end{array}\right.
$$
$\triangle$ Let's get rid of the denominator in the first equation:

$$
(x+y)(1+x y)=5 x y, \quad x+y+x y(x+y)=5 x y
$$

Now, it seems reasonable to divide all terms of the last equation by $x y$. But then we have to consider two cases.

1) Let $x y=0$.

If here $x=0$, then, as can be seen from the original system, $y=0, z=0$. Similarly, if $y=0$ or $z=0$, then the other two unknowns are zero. We get the solution $(0 ; 0 ; 0)$.

2) Let $x y \neq 0$.

After dividing the equation by $x y$ we have:

$$
\frac{1}{y}+\frac{1}{x}+x+y=5
$$

Similarly, the second and third equations of the original system can be transformed, under the condition $x y z \neq 0$, to the form:

$$
\frac{1}{z}+\frac{1}{y}+y+z=6, \quad \frac{1}{x}+\frac{1}{z}+z+x=7
$$

Introduce new variables:

$$
x+\frac{1}{x}=t, \quad y+\frac{1}{y}=u, \quad z+\frac{1}{z}=v
$$

We get a new system of equations:

$$
t+u=5, \quad u+v=6, \quad v+t=7
$$

From this, we find $t, u, v: t=3, u=2, v=4$. It remains to find $x, y$ and $z$.

Answer: $(0 ; 0 ; 0),((3 \pm \sqrt{5}) / 2 ; 1 ; 2 \pm \sqrt{3})$ (four solutions).