[tex]\mathbf f(x,y)=(2x-4y)\,\mathbf i+(-4x+10y-9)\,\mathbf j[/tex]
is conservative if
[tex]\dfrac{\partial(2x-4y)}{\partial y}=\dfrac{\partial(-4x+10y-9)}{\partial x}[/tex]
It's clear that this is true, since both partial derivatives reduce to -4, so [tex]\mathbf f[/tex] is indeed a conservative vector field.
Since [tex]\mathbf f(x,y)[/tex] is conservative, there must be some scalar function [tex]f(x,y)[/tex] such that
[tex]\nabla f(x,y)=\dfrac{\partial f}{\partial x}\,\mathbf i+\dfrac{\partial f}{\partial y}\,\mathbf j=\mathbf f(x,y)[/tex]
[tex]\implies\dfrac{\partial f}{\partial x}=2x-4y[/tex]
Integrating with respect to [tex]x[/tex] yields
[tex]\displaystyle\int\frac{\partial f}{\partial x}\,\mathrm dx=f(x,y)=x^2-4xy+g(y)[/tex]
and differentiating with respect to [tex]y[/tex] yields
[tex]\dfrac{\partial f}{\partial y}=-4x+g'(y)=-4x+10y-9[/tex]
[tex]\implies g'(y)=10y-9[/tex]
[tex]\implies g(y)=\displaystyle\int g'(y)\,\mathrm dy=5y^2-9y+C[/tex]
Thus the scalar potential of [tex]\mathbf f(x,y)[/tex] is
[tex]f(x,y)=x^2-4xy+5y^2-9y+C[/tex]
where [tex]C[/tex] is an arbitrary constant.
