Answer
$\|\mathbf{F}(x,y)\|=\frac{\sqrt{x^2+y^2}}{x^2+y^2}=\frac{1}{r},$
so vectors are very large near the origin (a singularity at $(0,0)$) and decay like $1/r$ as $r\to\infty$. In the sketch we omit the origin (field undefined there) and cap arrow lengths so the picture remains clear while preserving the relative change of vector size with distance.
Work Step by Step
The vector field is radial: at each point $(x,y)$, the vector points in the same direction as the position vector $\mathbf{r}=(x,y)$. Its magnitude is
$\|\mathbf{F}(x,y)\|=\frac{\sqrt{x^2+y^2}}{x^2+y^2}=\frac{1}{r},$
so vectors are very large near the origin (a singularity at $(0,0)$) and decay like $1/r$ as $r\to\infty$. In the sketch, we omit the origin (field undefined there) and cap arrow lengths so the picture remains clear while preserving the relative change of vector size with distance.
