Looks like we have
[tex]\hat F(x,y)=(y^2-4x)\,\hat\imath+2xy\,\hat\jmath[/tex]
[tex]\hat F(x,y)[/tex] is conservative if we can find a scalar function [tex]f(x,y)[/tex] such that [tex]\nabla f(x,y)=\hat F(x,y)[/tex]. For this to be the case, we would need to have
[tex]\dfrac{\partial f}{\partial x}=y^2-4x[/tex]
[tex]\dfrac{\partial f}{\partial y}=2xy[/tex]
Take the first partial differential equation. If we integrate both sides with respect to [tex]x[/tex], we'd get
[tex]\displaystyle\int\frac{\partial f}{\partial x}\,\mathrm dx=\int(y^2-4x)\,\mathrm dx[/tex]
[tex]\implies f(x,y)=xy^2-2x^2+g(y)[/tex]
Differentiating both sides with respect to [tex]y[/tex], we recover the other partial derivative and find
[tex]\dfrac{\partial f}{\partial y}=2xy+\dfrac{\mathrm dg}{\mathrm dy}=2xy\implies\dfrac{\mathrm dg}{\mathrm dy}=0\implies g(y)=C[/tex]
where [tex]C[/tex] is an arbitrary constant. So we've found
[tex]f(x,y)=xy^2-2x^2+C[/tex]
which means that [tex]\hat F(x,y)[/tex] is conservative.
