If $x^2-bx+c=(x+p)(x-q)$ ,then, factorise $x^2-bxy+cy^2$.
My attempt - $(x+p)(x-q)=x^2+px-qx-pq$
$\implies p-q=b $ and $pq=c$
similarly-$p'-q'=by$ and $p'q'=cy^2$(assuming that on factorising $x^2-bxy+cy^2$ the result is $(x+p')(x-q')$ ).What to do next?
Start with $$x^2-bx+c=(x+p)(x-q)$$ Now replace $x$ by $\frac{x}{y}$ $$\left(\frac{x}{y}\right)^2-b\frac{x}{y}+c=(\frac{x}{y}+p)(\frac{x}{y}-q)$$ now multiply by $y^2$ to get $$x^2-bxy+cy^2=(x+py)(x-qy)$$