Answer
$\mathbf{F}$ is conservative
Work Step by Step
Given
$$\mathbf{F}(x, y)= x y^2 \ \mathbf{i}+x^2 y \ \mathbf{j}$$
Since \begin{align}\mathbf{F}(x,y)&=M \mathbf{i}+N\mathbf{j}\\
&=f_{x}(x, y)\mathbf{i}+f_{y}(x, y)\mathbf{j}
\end{align}
we get
\begin{array}{l} {M=f_{x}(x, y)=\frac{\partial f(x,y)}{\partial x}=x y^2} \\ {N=f_{y}(x, y)=\frac{\partial f(x,y)}{\partial y}=x^2 y} \\ \end{array}
This implies that $M$ and $N$ have continuous first partial
derivatives.
Since
\begin{array}{l} { \frac{\partial M}{\partial y}=2 xy}
\\ { \frac{\partial N}{\partial x}=2xy} \\ \end{array}
So, we get
$$ \frac{\partial N}{\partial x}= \frac{\partial M}{\partial y}$$
and this implies that $\mathbf{F}$ is conservative