Answer
$V(x,y)= x^4y^5$
Work Step by Step
$\bf{Definition}:$ If a vector field $F=F_1 i+F_2j$ is a vector field over a simply connected and open set of domain $D$, then the partial derivatives of $ F_1$ and $ F_2$ must satisfy the condition such as: $\dfrac{\partial F_1}{\partial y}=\dfrac{\partial F_2}{\partial x}$
We are given that $F(x,y)=(4x^3y^5, 5x^4y^4)$
$\dfrac{\partial F_1}{\partial y}=20x^3y^4 \quad \text{and}\quad
\dfrac{\partial F_2}{\partial x}=20x^3y^4 $
We see that the partial derivative of a given vector field are equal. Therefore, the vector field $F$ is conservative.
Now, the potential function is: $V(x,y)=\int 4x^3y^5 dx= x^4y^5$
