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Take $E$ to be continuously differentiable everywhere on ${\mathbb{R}}^3$, as in the question. Its zero curl says ${\partial }_i{E}_j={\partial }_j{E}_i$. Construct a potential of a conservative vector field explicitly byAll segments lie in ${\mathbb{R}}^3$. Differentiating under the integral and using the symmetry of the derivatives,Hence $E=-\nabla \varphi$. This radial integral potential for a curl-free field also works on a star-shaped domain after translating its centre. On an arbitrary domain, zero curl only guarantees local potentials; the everywhere-on-${\mathbb{R}}^3$ interpretation is important for this global assertion.

For the given field, put $q=e^{-x^{2}z}$. The three pairs of derivatives entering its curl areThus the field is irrotational. To find its potential using $E_y=-{\partial }_y\varphi$, integrate ${\partial }_y\varphi =2yq$ to get $\varphi =y^{2}q+\psi (x,z)$. Comparison with $E_x=-{\partial }_x\varphi$ gives ${\psi }_x=0$, and comparison with $E_z=-{\partial }_z\varphi$ gives ${\psi }_z=0$. ThereforeDifferentiating this expression reproduces all three components with the required minus sign; the only freedom on the connected domain is the additive constant.

The sufficiency claim requires a domain qualification. For a continuously differentiable vector field on all of ${\mathbb{R}}^3$, or more generally on a simply connected open domain, the equivalence is valid. On an arbitrary domain, vanishing curl is necessary but may fail to be sufficient.

Necessity. Suppose every closed-curve line integral vanishes. Fix a point $p$ in the open domain and a unit vector $a$. A sufficiently small planar disk $D_{\epsilon }$ of radius $\epsilon$, centred at $p$ and normal to $a$, lies in the domain. Stokes theorem givesDivide by $\pi {\epsilon }^2$ and let $\epsilon \downarrow 0$. By continuity, the average tends to $(\nabla \times \mathbf{F})(p)\cdot a$. Taking the three coordinate directions proves $\boxed{\nabla \times \mathbf{F}=0}$ everywhere.

Sufficiency on ${\mathbb{R}}^3$. A direct proof constructs a potential of a conservative vector field. PutWrite $F_j$ for the components of $\mathbf{F}$. Vanishing curl says ${\partial }_i{F}_j={\partial }_j{F}_i$. Differentiation under the integral, followed by the chain rule, givesThis is the radial integral potential for a curl-free field. The fundamental theorem for line integrals now gives zero on every closed curve. On a simply connected open domain the analogous argument uses path integrals: every loop can be contracted within the domain, and applying Stokes theorem to a piecewise smooth contraction makes its circulation zero. Equivalently, whenever a particular loop bounds an oriented spanning surface entirely in the domain, Stokes theorem immediately gives zero for that loop.

Why the domain matters. On $U={\mathbb{R}}^3\setminus \{(0,0,z):z\in \mathbb{R}\}$, the smooth vector fieldhas zero curl, but its line integral around the unit circle in $z=0$ is $2\pi$, as in ii. No spanning surface for this loop can lie entirely in $U$. The literal claim without a condition on the domain therefore has this counterexample.