= Solution
Take $E$ to be continuously differentiable everywhere on $\mathbb R^3$, as in the question. Its zero <curl> says $\partial_iE_j=\partial_jE_i$. Construct a <potential of a conservative vector field> explicitly by
$$
\phi(x)=-\int_0^1E(tx)\cdot x\,dt.
$$
All segments lie in $\mathbb R^3$. Differentiating under the integral and using the symmetry of the derivatives,
$$
\partial_i\phi=-\int_0^1[E_i(tx)+t x_j\partial_iE_j(tx)]\,dt=-\int_0^1[E_i(tx)+t x_j\partial_jE_i(tx)]\,dt=-\int_0^1\frac{d}{dt}[tE_i(tx)]\,dt=-E_i(x).
$$
Hence \b[$E=-\nabla\phi$]. 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 $q=e^{-x^2z}$. The three pairs of derivatives entering its <curl> are
$$
\partial_yE_z=2x^2yq=\partial_zE_y,\qquad \partial_zE_x=2xy^2(1-x^2z)q=\partial_xE_z,\qquad \partial_xE_y=4xyzq=\partial_yE_x.
$$
Thus the field is <irrotational>. To find its potential using $E_y=-\partial_y\phi$, integrate $\partial_y\phi=2yq$ to get $\phi=y^2q+\psi(x,z)$. Comparison with $E_x=-\partial_x\phi$ gives $\psi_x=0$, and comparison with $E_z=-\partial_z\phi$ gives $\psi_z=0$. Therefore
$$
\boxed{\phi(x,y,z)=y^2e^{-x^2z}+C}.
$$
Differentiating this expression reproduces all three components with the required minus sign; the only freedom on the connected domain is the additive constant.
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