Verified answer

[(x²-xy-20y²) / (x²-8xy+15y²)] / [(x²+2xy-8y²) / (x²-xy-6y²)]

= [(x-5y)(x+4y)/((x-5y)(x-3y))] / [(x+4y)(x-2y) / ((x+2y)(x-3y))]

= [(x+4y)/(x-3y)] * [((x+2y)(x-3y)) / ((x+4y)(x-2y))]

= (x+2y) / (x-2y)

the first thing you should recognize is that when you divide a fraction by a fraction, you multiply the first one by the reciprocal of the second

[(x^2 - xy - 20y^2)(x^2 - xy - 6y^2)]/[(x^2 - 8xy + 15y^2)(x^2 + 2xy - 8y^2)]

factor all of your quadratics

[(x-5y)(x+4y)(x-3y)(x+2y)]/[(x-3y)(x-5y)(x+4y)(x-2y)]

You can then cancel out like terms from the numerator and denominator. You are left with:

(x+2y)/(x-2y)

-2+2.5y^2/x^2-2.5y^2

that's everything i know