Answer
$$\mathbf{F} \ \text {is conservative}$$ $$f(x, y)= \frac{1}{2}\ln (x^{2} +y^2) +C $$
Work Step by Step
Given $$\mathbf{F}(x, y)=\frac{x \mathbf{i}+y \mathbf{j}}{x^{2}+y^{2}}=\frac{x }{x^{2}+y^{2}}\mathbf{i}+\frac{y }{x^{2}+y^{2}}\mathbf{j}$$ Since we have
\begin{align}\mathbf{F}(x,y)&=M \mathbf{i}+N\mathbf{j}\end{align}
So, we get \begin{array}{l} {M=\frac{x }{x^{2}+y^{2}} } \\ {N= \frac{y }{x^{2}+y^{2}}} \end{array} this implies that $M$ and $N$ have continuous first partial derivatives at $x, y\neq 0$. As, we have \begin{array}{l} {\begin{align} \frac{\partial M}{\partial y}&=\frac{0-2xy }{(x^{2}+y^{2})^2}\\ &=-\frac{2xy }{(x^{2}+y^{2})^2} \end{align}} \\ {\begin{aligned} \frac{\partial N}{\partial x}&=\frac{0-2xy }{(x^{2}+y^{2})^2}\\ &=-\frac{2xy }{(x^{2}+y^{2})^2} \end{aligned}} \\ \end{array} So, we get $$ \frac{\partial N}{\partial x}= \frac{\partial M}{\partial y}$$ and this implies that $\mathbf{F}$ is conservative. By definition, If a vector field $\mathbf{F}$ is conservative, then there exists a function $f$, Such that \begin{align} {\qquad \mathbf{F}(x,y)=\nabla f(x,y)=f_{x}(x,y) \mathbf{i}+f_{y}(x,y) \mathbf{j}} \end{align} so, we get \begin{array}{l} { f_{x}(x, y)=\frac{ x }{x^{2}+y^{2}} \ \ \ \ \mathbf{\rightarrow (1)}} \\ { f_{y}(x, y)= \frac{y }{x^{2}+y^{2}} \ \ \ \mathbf{\rightarrow (2)}} \\ \end{array} ${\text { Integrate both sides of Eq.(1) with respect to } x}\\{ \text { assuming } y \text { as constant, so we get }} $ \begin{array}{l} { f(x,y)= \int f_{x}(x, y) dx= \frac{1}{2} \int \frac{2x }{x^{2}+y^{2}} dx\\ = \frac{1}{2}\ln (x^{2} +y^2) +g(y) \mathbf{\rightarrow (3)}} \end{array} ${\text { Partially differentiate } Eq.(3) \text { with respect to } y, \text { To get }}$ $$ f_{y}(x, y)=\frac{y }{x^{2}+y^{2}} +g^{\prime}(y) \mathbf{\rightarrow (4)}$$ Compare $Eq.(4)$ with $Eq.(2)$, To get \begin{array}{l} {\qquad g^{\prime}(y)=0} \\ { \Rightarrow \ \ \ g(y)=C} \end{array} Substitute this in $Eq.(3)$, we get $$f(x, y)= \frac{1}{2}\ln (x^{2} +y^2) +C $$
