(a) $\mathbf{F}=2xy\mathbf{i}+(x^{2}+1)\mathbf{j}$ is a conservative vector field on the entire $xy$-plane since $$\begin{equation*} \dfrac{\partial P}{\partial y}=\dfrac{\partial }{\partial y}(2xy)=2x\qquad \hbox{and}\qquad \dfrac{\partial Q}{\partial x}=\dfrac{\partial }{\partial x}(x^{2}+1)=2x \end{equation*}$$

are equal for any choice of $(x,y)$.

(b) The vector field $\mathbf{F}=\dfrac{x}{y^{2}}\mathbf{i}-\dfrac{x^{2}}{y^{3}}\mathbf{j}$ is conservative on any connected region not containing points on the $x$-axis $(y=0)$ since $$\begin{equation*} \dfrac{\partial P}{\partial y}=\dfrac{\partial }{\partial y}\dfrac{x}{y^{2}} =-\dfrac{2x}{y^{3}} \qquad \hbox{and}\qquad \dfrac{\partial Q}{\partial x}=\dfrac{\partial }{\partial x}\left( -\dfrac{x^{2}}{y^{3}}\right) =-\dfrac{2x}{y^{3}} \end{equation*}$$

are equal, provided $y\neq 0$. Because $\mathbf{F}=\dfrac{x}{y^{2}}\mathbf{i}-\dfrac{x^{2}}{y^{3}}\mathbf{j}$ is conservative for $y\neq 0,$ the line integral $\int_{C}\mathbf{F}\,{\cdot}\, d\mathbf{r}=\int_{C}\left[ \dfrac{x}{y^{2}}dx-\dfrac{x^{2}}{y^{3}}dy\right]$ is independent of the path in any simply connected region not containing points on the $x$-axis.