(A)
[tex]\dfrac{\partial f}{\partial x}=-16x+2y[/tex]
[tex]\implies f(x,y)=-8x^2+2xy+g(y)[/tex]
[tex]\implies\dfrac{\partial f}{\partial y}=2x+\dfrac{\mathrm dg}{\mathrm dy}=2x+10y[/tex]
[tex]\implies\dfrac{\mathrm dg}{\mathrm dy}=10y[/tex]
[tex]\implies g(y)=5y^2+C[/tex]
[tex]\implies f(x,y)=\boxed{-8x^2+2xy+5y^2+C}[/tex]
(B)
[tex]\dfrac{\partial f}{\partial x}=-8y[/tex]
[tex]\implies f(x,y)=-8xy+g(y)[/tex]
[tex]\implies\dfrac{\partial f}{\partial y}=-8x+\dfrac{\mathrm dg}{\mathrm dy}=-7x[/tex]
[tex]\implies \dfrac{\mathrm dg}{\mathrm dy}=x[/tex]
But we assume [tex]g(y)[/tex] is a function of [tex]y[/tex] alone, so there is not potential function here.
(C)
[tex]\dfrac{\partial f}{\partial x}=-8\sin y[/tex]
[tex]\implies f(x,y)=-8x\sin y+g(x,y)[/tex]
[tex]\implies\dfrac{\partial f}{\partial y}=-8x\cos y+\dfrac{\mathrm dg}{\mathrm dy}=4y-8x\cos y[/tex]
[tex]\implies\dfrac{\mathrm dg}{\mathrm dy}=4y[/tex]
[tex]\implies g(y)=2y^2+C[/tex]
[tex]\implies f(x,y)=\boxed{-8x\sin y+2y^2+C}[/tex]
For (A) and (C), we have [tex]f(0,0)=0[/tex], which makes [tex]C=0[/tex] for both.