Given,

Line $6x^2-xy-12y^2=0$

$\Rightarrow 6x^2-9xy+8xy-12y^2=0$

$\Rightarrow \left(3x+4y\right)\left(2x-3y\right)=0 ..........\left(1\right)$

Now solving $15x^2+14xy-8y^2=0$ we get,

$\Rightarrow 15x^2+20xy-6xy-8y^2=0$

$\Rightarrow \left(5x-2y\right)\left(3x+4y\right)=0 ......\left(2\right)$

From equation $\left(1\right) \& \left(2\right)$ we get common line as $3x+4y=0$

Now line passing through the point $\left(-1,1\right)$ and parallel to the common line will be $3x+4y-1=0$

Now we will find the equation of pair of lines joining the origin to the points of intersection of the curve $2x^2-xy-y^2+x-y=0$ and the line $3x+4y-1=0$ by homogenising the equations,

So putting the value of $1=3x+4y$ in $2x^2-xy-y^2+x\times 1-y\times 1=0$ we get,

$\Rightarrow 2x^2-xy-y^2+x\times \left(3x+4y\right)-y\times \left(3x+4y\right)=0$

$\Rightarrow 2x^2-xy-y^2+3x^2+4xy-3xy-4y^2=0$

$\Rightarrow 2x^2-y^2+3x^2-4y^2=0$

$\Rightarrow 5x^2-5y^2=0$

$\Rightarrow x^2-y^2=0$