6.072. $\left\{\begin{array}{l}\frac{1}{y-1}-\frac{1}{y+1}=\frac{1}{x}, \\ y^{2}-x-5=0 .\end{array}\right.$
Solution.

Domain of definition: $\left\{\begin{array}{l}y \neq \pm 1, \\ x \neq 0 .\end{array}\right.$

Transform the first equation of the system

$$
\left\{\begin{array} { l } 
{ x ( y + 1 ) - x ( y - 1 ) = y ^ { 2 } - 1 , } \\
{ y ^ { 2 } - x = 5 }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ x y + x - x y + x = y ^ { 2 } - 1 , } \\
{ y ^ { 2 } - x = 5 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
2 x=y^{2}-1, \\
y^{2}-x=5
\end{array} \Rightarrow\right.\right.\right.
$$

$\Rightarrow x=y^{2}-5$. Substituting this value of $x$ into the first equation of the system, we get $2\left(y^{2}-5\right)=y^{2}-1, y^{2}=9, y_{1}=3, y_{2}=-3$; then $x_{1}=4, x_{2}=4$. Answer: $(4 ; 3)(4 ;-3)$.