Answer
Conservative
$f(x,y)=xy^2-x^2+C$
Work Step by Step
When $F(x,y)=pi+qj$ is a conservative field, then throughout the $D$, then we have
$\dfrac{\partial p}{\partial y}=\dfrac{\partial q}{\partial x}$
Here, $p$ and $q$ represents the first-order partial derivatives on a domain $D$.
As we are given that
$F(x,y)=(y^2-2x)i+2xy)$
Then, we have $p_x=2y; p_y=2y$
Here, $\dfrac{\partial p}{\partial y} = \dfrac{\partial q}{\partial x}$
so, the $F$ is conservative.
Need to find the function $f$ such that $F=\nabla f$
The partial derivatives of $f$ are: $f_x=y^2-2x..(a) \\ f_y=2xy ...(b)$
integrate $f_x$ with respect to $x$.
$f(x,y)=xy^2-x^2+g(y)$ ...(c)
Here, $g(y)$ is a constant of integration with respect to $x$ and with a function of $y$.
Take derivative of the equation $c$ with respect to $y$.
$f_y(x,y)=2xy+g'(y) ...(d)$
On comparing the equations (b) and (d) , we get
$f_y(x,y)=2xy$ and $g'(y)=0$
Now, $g(y)=C$ [ Integrate with respect to $y$]
Here, $C$ is a constant of integration.
Hence, $f(x,y)=xy^2-x^2+C$
