Answer
$\mathbf{F}$ is $\mathbf{not}$ conservative
Work Step by Step
Given
$$\mathbf{F}(x, y)=\frac{1}{\sqrt{x^{2}+y^{2}}}(\mathbf{i}+\mathbf{j}) =\frac{1}{\sqrt{x^{2}+y^{2}}} \mathbf{i} \frac{1}{\sqrt{x^{2}+y^{2}}} \mathbf{j} $$
Since \begin{align}\mathbf{F}(x,y)&=M \mathbf{i}+N\mathbf{j}\\
&=f_{x}(x, y)\mathbf{i}+f_{y}(x, y)\mathbf{j}
\end{align}
We get
\begin{array}{l}
{M=f_{x}(x, y)=\frac{\partial f(x,y)}{\partial x}=\frac{1}{\sqrt{x^{2}+y^{2}}} =(x^{2}+y^{2})^{-\frac{1}{2}}} \\
{N=f_{y}(x, y)=\frac{\partial f(x,y)}{\partial y}=\frac{1}{\sqrt{x^{2}+y^{2}}}=(x^{2}+y^{2})^{-\frac{1}{2}}} \\ \end{array}
This implies that $M$ and $N$ have continuous first partial
derivatives.
Since
\begin{array}{l} { \frac{\partial M}{\partial y}=-\frac{1}{2}(2y)(x^{2}+y^{2})^{-\frac{3}{2}}=-y(x^{2}+y^{2})^{-\frac{3}{2}} }
\\ { \frac{\partial N}{\partial x}=-\frac{1}{2}(2x)(x^{2}+y^{2})^{-\frac{3}{2}}=-x(x^{2}+y^{2})^{-\frac{3}{2}}} \\ \end{array}
So, we get
$$ \frac{\partial N}{\partial x}\ne \frac{\partial M}{\partial y}$$
And this implies that $\mathbf{F}$ is $\mathbf{not}$ conservative