If there is a scalar function [tex]f(x,y)[/tex] such that [tex]\mathbf f(x,y)=\nabla f(x,y)[/tex], then it would satisfy
[tex]\dfrac{\partial f}{\partial x}=x^2y^3[/tex]
[tex]\dfrac{\partial f}{\partial y}=x^3y^2[/tex]
From the first PDE, we have
[tex]f(x,y)=\dfrac13x^3y^3+g(y)[/tex]
Take the derivative with respect to [tex]y[/tex] to get
[tex]\dfrac{\partial f}{\partial y}=x^3y^2+\dfrac{\mathrm dg}{\mathrm dy}=x^3y^2[/tex]
[tex]\implies\dfrac{\mathrm dg}{\mathrm dy}=0\implies g(y)=C[/tex]
So,
[tex]f(x,y)=\dfrac13x^3y^3+C[/tex]
By the fundamental theorem of calculus,
[tex]\displaystyle\int_{\mathcal C}\mathbf f(x,y)\cdot\mathrm d\mathbf r=\int_{\mathcal C}\nabla f(x,y)\cdot\mathrm d\mathbf r=f(\mathbf r(1))-f(\mathbf r(0))[/tex]
[tex]=f(-1,3)-f(0,0)=-9[/tex]
