Answer
a) $\dfrac{\partial P}{\partial y} = \dfrac{\partial Q}{\partial x}$
b) $\int_C \overrightarrow{F} \cdot \overrightarrow{dr}$ is not independent of path.
Work Step by Step
When $F(x,y)=Pi+Qj$ is a conservative field, then throughout the domain $D$, we get
$\dfrac{\partial P}{\partial y}=\dfrac{\partial Q}{\partial x}$
$a$ and $b$ are the first-order partial derivatives on the domain $D$.
Here, we have $\dfrac{\partial P}{\partial y} = \dfrac{\partial Q}{\partial x}=\dfrac{y^2-x^2}{(x^2+y^2)^2}$
Thus, the vector field $F$ is conservative.
Hence, $\dfrac{\partial P}{\partial y} = \dfrac{\partial Q}{\partial x}$
b) The line integral $\int_C \overrightarrow{F} \cdot \overrightarrow{dr}$ is independent of the path if and only if $\int_C \overrightarrow{F} \cdot \overrightarrow{dr}=0$ for every closed curve $C$.
Here, $\int_{C_1} \overrightarrow{F} \cdot \overrightarrow{dr}=\int_0^{\pi} (-\sin \theta i+\cos \theta j) \cdot (-\sin \theta i+\cos \theta j) d\theta= \int_0^{\pi} 1 d\theta =\pi$
and
$\int_{C_2} \overrightarrow{F} \cdot \overrightarrow{dr}=\int_0^{-\pi} (-\sin \theta i+\cos \theta j) \cdot (-\sin \theta i+\cos \theta j) d\theta= \int_0^{-\pi} 1 d\theta =-\pi$
Here, the two integrals differ, so the line integral $\int_C \overrightarrow{F} \cdot \overrightarrow{dr}$ is not independent of path.
