= Solution
For $r>0$, the <chain rule> gives $\partial r/\partial x_i=x_i/r$. Differentiating the radial factor gives $d\phi/dr=e^{ikr}(ikr-1)/(4\pi r^2)$, so in <Cartesian coordinates>
$$
\boxed{\nabla\phi=\frac{e^{ikr}(ikr-1)}{4\pi r^3}(x,y,z).}
$$
The origin is excluded because the scalar field is singular there.
The <Stokes theorem> states $\int_S(\nabla\times\mathbf F)\cdot\mathbf n\,dS=\oint_{\partial S}\mathbf F\cdot d\boldsymbol\ell$ for a sufficiently smooth vector field and oriented surface with piecewise smooth boundary. The boundary orientation follows the right-hand rule relative to $\mathbf n$.
For the <convective acceleration identity>, use the <Levi-Civita symbol> contraction $\epsilon_{ijk}\epsilon_{k\ell m}=\delta_{i\ell}\delta_{jm}-\delta_{im}\delta_{j\ell}$. It gives
$$
[\mathbf F\times(\nabla\times\mathbf F)]_i=\epsilon_{ijk}F_j\epsilon_{k\ell m}\partial_\ell F_m=F_j\partial_iF_j-F_j\partial_jF_i=\frac12\partial_i(F_jF_j)-[(\mathbf F\cdot\nabla)\mathbf F]_i.
$$
Rearranging proves the required vector identity.
For <orthogonal coordinates>, define the <scale factors of orthogonal coordinates> $h_i=|\partial\mathbf r/\partial q_i|$ and tangent <unit vectors> $\mathbf e_i=h_i^{-1}\partial\mathbf r/\partial q_i$. They form an <orthonormal basis>, and $d\mathbf r=\sum_i h_i\mathbf e_i\,dq_i$. Equating $d\psi=\sum_i\partial_{q_i}\psi\,dq_i$ with $\nabla\psi\cdot d\mathbf r$ identifies each component. Thus the <gradient in orthogonal coordinates> is
$$
\boxed{\nabla\psi=\sum_{i=1}^3\frac1{h_i}\frac{\partial\psi}{\partial q_i}\mathbf e_i.}
$$
This holds on a regular coordinate chart where all scale factors are positive.
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