The lines represented by the equation $x^2-y^2-x+3y-2=0$ are

$x^2-y^2-x+3y-2=0\to \left(1\right)$

$$\begin{gathered} Now x^2-y^2=(x+y)(x-y) \\ let the lines represented by \left(1\right) are \\ x+y+l=0 and x-y+k=0 \\ \therefore x^2-y^2-x+3y-2=(x+y+l)(x-y+k) \\ multiply and compare like terms on both sides we get l=-2,k=1 \end{gathered}$$

$\therefore$The required lines are $x-y+1=0, x+y-2=0$