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
The vector field $F(x,y)=Pi+Qj$ is known as conservative field throughout the domain $D$, when we have
$\dfrac{\partial P}{\partial y}=\dfrac{\partial Q}{\partial x}$
$P$ and $Q$ represents the first-order partial derivatives on the domain $D$.
From the given problem, we get $\dfrac{\partial P}{\partial y}=\dfrac{\partial Q}{\partial x}=\dfrac{y^2-x^2}{(x^2+y^2)^2}$
Thus, we have the given the vector field $F$ a conservative field.
Hence, we get $\dfrac{\partial P}{\partial y} = \dfrac{\partial Q}{\partial x}$
b) The line integral $\int_C \overrightarrow{F} \cdot \overrightarrow{dr}$ is independent of path when $\int_C \overrightarrow{F} \cdot \overrightarrow{dr}=0$ for every closed curve $C$.
Here, we have $\int_{C_1} \overrightarrow{F} \cdot \overrightarrow{dr}=\int_0^{\pi} (-\sin \theta i+\cos \theta j) (-\sin \theta i+\cos \theta j) d\theta$
and $ \int_0^{\pi} 1 d\theta =\pi$
Also, $\int_{C_2} \overrightarrow{F} \cdot \overrightarrow{dr}=\int_0^{-\pi} (-\sin \theta i+\cos \theta j) (-\sin \theta i+\cos \theta j) d\theta$
and $ \int_0^{-\pi} 1 d\theta =-\pi$
It has been seen that the two integrals are differ, thus the line integral $\int_C \overrightarrow{F} \cdot \overrightarrow{dr}$ is not independent of path.
