Let $\frac{s}{r}$ be a vector field such that: $s=-yi+xj,r=\sqrt{x^2+y^2}$
Is $\frac{s}{r}$ a conservative field?
My attempt:
$\frac{s}{r}$ is a conservative field $\iff \displaystyle\oint F\cdot dr=0$
$$\frac{s}{r}=\frac{-y\hat i+x\hat j}{\sqrt{x^2+y^2}}$$
$$\boxed{=\frac{-y}{\sqrt{x^2+y^2}}\cdot \hat i+\frac{x}{\sqrt{x^2+y^2}} \cdot \hat j }$$
Let $Q:=\frac{x}{\sqrt{x^2+y^2}}$ , let $P:=\frac{-y}{\sqrt{x^2+y^2}} $
$\frac{s}{r}$ is a conservative field $\iff \frac{\partial Q}{\partial x}=\frac{\partial P}{\partial x}$
Finding $\frac{\partial Q}{\partial x}$: $\color{blue}{\frac{\partial Q}{\partial x}=\frac{y^2}{(x^2+y)^{3/2}}}$
Finding $\frac{\partial P}{\partial x}$: $\color{blue}{\frac{\partial P}{\partial y}=-\frac{y^2}{(x^2+y)^{3/2}}}$
$\Longrightarrow \frac{\partial P}{\partial x} \neq \frac{\partial P}{\partial x}$ therfore $\frac{s}{r}$ is not conservative field.
Is it correct?
The usual way to show that a particular vector field is not conservative is to take the line integral around a closed loop and see that the result is not zero.
For your field, taking the line integral around the unit circle gives a value of $2\pi$, so the field is not conservative. In fact, your field and that line integral are (almost) the examples given by Wikipedia for a non-conservative field.