Assuming the field is given by
[tex]\mathbf f(x,y)=ye^{xy}\,\mathbf i+xe^{xy}\,\mathbf j[/tex]
then because [tex]\mathbf f(x,y)[/tex] is continuous, there is some scalar potential function [tex]f(x,y)[/tex] such that [tex]\nabla f(x,y)=\mathbf f(x,y)[/tex], i.e. the vector field is conservative. So
[tex]\dfrac{\partial f(x,y)}{\partial x}=ye^{xy}[/tex]
[tex]\displaystyle\int\dfrac{\partial f}{\partial x}=\dfrac{ye^{xy}}y+g(y)=e^{xy}+g(y)[/tex]
[tex]\dfrac{\partial f(x,y)}{\partial y}=\dfrac{\partial(e^{xy}+g(y))}{\partial y}[/tex]
[tex]xe^{xy}=xe^{xy}+g'(y)[/tex]
[tex]0=g'(y)[/tex]
[tex]\implies g(y)=C[/tex]
[tex]\implies f(x,y)=e^{xy}+C[/tex]
The value of the line integral then depends on only the endpoints of the path and the gradient theorem (fundamental theorem of calculus for line integrals) applies.
