Answer
$\mathbf{F}$ is conservative
Work Step by Step
Given
$$\mathbf{F}(x, y)=\frac{1}{\sqrt{1+xy }}(y\mathbf{i}+x\mathbf{j}) =\frac{y}{\sqrt{ 1+xy}} \mathbf{i} +\frac{x}{\sqrt{1+xy}} \mathbf{j} $$
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{y}{\sqrt{1+xy}}=y(1+xy)^{-\frac{1}{2}}} \\
{N=f_{y}(x, y)=\frac{\partial f(x,y)}{\partial y}=\frac{x}{\sqrt{1+xy}}=x(1+xy)^{-\frac{1}{2}}} \\ \end{array}
This implies that $M$ and $N$ have continuous first partial
derivatives.
Since
\begin{array}{l} {\begin{align} \frac{\partial M}{\partial y}&=(1+xy)^{-\frac{1}{2}}+y (-\frac{1}{2})x(1+xy)^{-\frac{3}{2}}\\
&=(1+xy)^{-\frac{1}{2}}+-\frac{1}{2}xy(1+xy)^{-\frac{3}{2}}\\
&=\frac{1}{2}(1+xy)^{-\frac{3}{2}}(2+2xy-xy)\\
&= (1+xy)^{-\frac{3}{2}}(2+ xy )
\end{align}}
\\ {\begin{aligned} \frac{\partial N}{\partial x}&=(1+xy)^{-\frac{1}{2}}+x (-\frac{1}{2})y(1+xy)^{-\frac{3}{2}}\\
&=\frac{1}{2}(1+xy)^{-\frac{3}{2}}(2+2xy-xy)\\
&= (1+xy)^{-\frac{3}{2}}(2+ xy )
\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