Answer
$\phi(x,y)=\dfrac{1}{2}\ln (x^2+y^2)$
Work Step by Step
For a vector field to be Conservative, $\dfrac{\partial f}{\partial y}=\dfrac{\partial g}{\partial x}$
We have: $f(x,y)=\dfrac{x}{x^2+y^2}$ and $g(x,y)=\dfrac{y}{x^2+y^2}$
Thus, $\dfrac{\partial f}{\partial y}=\dfrac{2xy}{(x^2+y^2)^2}$ and $\dfrac{\partial g}{\partial x}=\dfrac{2xy}{(x^2+y^2)^2}$
Therefore, a vector field $F$ is Conservative.
Now, potential function $F=\nabla \phi$
So, $\dfrac{\partial \phi}{\partial x}=\dfrac{x}{x^2+y^2} $ and $\phi(x,y)=\dfrac{1}{2}\ln (x^2+y^2) +h(y)$
and $\dfrac{\partial \phi}{\partial y}=\dfrac{y}{x^2+y^2}+h^{\prime}(y)$ and $h(y)=C(y)$
Thus, $\phi(x,y)=\dfrac{1}{2}\ln (x^2+y^2)$
