 Solution

Integrating the first component in $x$ gives $\varphi =x^{3}y+h(y,z)$. Matching the second requires $h_y=-2y{z}^2$, so $h=-y^2{z}^2+k(z)$. Matching the third gives $k^{\prime }=0$. Thus the field is conservative, with scalar potentialDirect differentiation verifies ${\mathbf{F}}_1=\nabla \varphi$ on all of ${\mathbb{R}}^3$.