We need to have
[tex]\dfrac{\partial f}{\partial x}=2x-4y[/tex]
[tex]\dfrac{\partial f}{\partial y}=-4x+10y-5[/tex]
Integrate the first PDE with respect to [tex]x[/tex]:
[tex]\displaystyle\int\frac{\partial f}{\partial x}\,\mathrm dx=\int(2x-4y)\,\mathrm dx\implies f(x,y)=x^2-4xy+g(y)[/tex]
Differentiate with respect to [tex]y[/tex] to get
[tex]\dfrac{\partial f}{\partial y}=-4x+10y-5=-4x+\dfrac{\mathrm dg}{\mathrm dy}\implies\dfrac{\mathrm dg}{\mathrm dy}=10y-5[/tex]
[tex]\implies g(y)=5y^2-5y+C[/tex]
So we have
[tex]f(x,y)=x^2-4xy+5y^2-5y+C[/tex]
which means [tex]F[/tex] is indeed conservative.
