Answer
$\phi(x,y,z)=\dfrac{1}{2}\ln (x^2+y^2+z^2)$
Work Step by Step
For a vector field to be Conservative, $\dfrac{\partial f}{\partial y}=\dfrac{\partial g}{\partial x}$ and $\dfrac{\partial g}{\partial z}=\dfrac{\partial h}{\partial y}$ and $\dfrac{\partial f}{\partial z}=\dfrac{\partial h}{\partial x}$
We have: $f(x,y,z)=\dfrac{x}{x^2+y^2+z^2}, g(x,y,z) =\dfrac{y}{x^2+y^2+z^2}$ and $h(x,y,z)=\dfrac{z}{x^2+y^2+z^2}$
Thus, $\dfrac{\partial f}{\partial y}=\dfrac{2xy}{(x^2+y^2+z^2)^2}=\dfrac{\partial g}{\partial x}$ and $\dfrac{\partial g}{\partial z}=\dfrac{2yz}{(x^2+y^2+z^2)^2}=\dfrac{\partial h}{\partial y}$ and $\dfrac{\partial f}{\partial z}=\dfrac{2xz}{(x^2+y^2+z^2)^2}=\dfrac{\partial h}{\partial x}$
Therefore, a vector field $F$ is Conservative.
Now, potential function $F=\nabla \phi$
So, $\dfrac{\partial \phi}{\partial x}=\dfrac{x}{(x^2+y^2+z^2)^2} \implies \phi(x,y,z)=\dfrac{1}{2} \ln (x^2+y^2+z^2)+a(y,z)$
and $\dfrac{\partial \phi}{\partial y}=\dfrac{y}{(x^2+y^2+z^2)^2} =\dfrac{y}{(x^2+y^2+z^2)^2} +\dfrac{\partial a}{\partial y} \implies \dfrac{\partial a}{\partial y}=0 \implies a(y,z)=b(z)$ and $\dfrac{\partial \phi}{\partial z}=\dfrac{y}{(x^2+y^2+z^2)^2} =\dfrac{y}{(x^2+y^2+z^2)^2} +b'(z) \implies b(z)=c$
Thus, $\phi(x,y,z)=\dfrac{1}{2}\ln (x^2+y^2+z^2)$
