Answer
Conservative
$f(x,y)=yxe^{xy}+k$
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 $D$.
We are given that $F(x,y)=y^2e^{xy}i+(1+xy)e^{xy}j$
Here, $p_x=2ye^{xy}+xy^2e^{xy}; q_y=2ye^{xy}+xy^2e^{xy}$
Therefore, we can see that $\dfrac{\partial p}{\partial y} = \dfrac{\partial q}{\partial x}$
Thus, the $F$ is conservative.
Need to find the function $f$ such that $F=\nabla f$
The partial derivatives of $f$ are:
$f_x=y^2e^{xy}..(a) \\ f_y=(1+xy)e^{xy} ...(b)$
Integrate $f_x$ with respect to $x$.
$f(x,y)=ye^{xy}+g(y)$ ...(c)
Here, $g(y)$ is a constant of integration.
Take derivative of the equation $c$ with respect to $y$.
we have $f_y(x,y)=yxe^{xy}+e^{xy}+g'(y) ...(d)$
On comparing the equations (b) and (d) , we get
$g'(y)=0$
Now, integrate it with respect to $y$.
This implies that $g(y)=C$
Here, $C$ is a constant of integration.
Hence, $f(x,y)=yxe^{xy}+k$
