My textbook says the following vector field is radial. How shall I check it? $F(x,y)=(x^2,xy)$
Let us take three different radius of same magnitude and check whether the magnitude of vector at these points (equidistant from the centre) are same.
Points $(10,0),(6,8),(0,10)$ are equidistant from the centre. The vectors at these points are $(100,0),(36,48),(0,0)$. But their magnitudes are not same. Then how can this be a radial vector field?
A vector field $F(x,y)$ is radial if each vector in the field points either towards or away from the origin that is iff is $F(x,y)$ is proportional to the position vector $(x,y)$.
In your case $F(x,y)=x(x,y)$. It is zero along the line $x=0$ and it points away from the origin elsewhere. Note that this vector field is not conservative in $\mathbb{R}^2$ because for $y\not=0$, $$y=\frac{\partial F_y}{\partial x} \not=\frac{\partial F_x}{\partial y}=0.$$