= Solution
A continuously differentiable <vector field> is an <irrotational vector field> when its <curl> vanishes:
$$
\nabla\times\mathbf F=\mathbf0.
$$
For the global-potential construction, assume that the domain is connected and simply connected, or more generally that all closed-curve <line integrals> of $\mathbf F$ vanish. Then define the <potential of a conservative vector field> by
$$
\boxed{V(\mathbf x)=-\int_{\mathbf x_0}^{\mathbf x}\mathbf F\cdot d\mathbf r.}
$$
Two paths with the same endpoints differ by a closed curve. On a <simply connected> domain that curve can be spanned, or contracted through piecewise smooth spanning surfaces, inside the domain. <Stokes theorem> gives zero circulation because the <curl> is zero. The value of $V$ is therefore path-independent and $V(\mathbf x_0)=0$. To compute a derivative, append a short coordinate segment to a path ending at $\mathbf x$. Then $V(\mathbf x+h\mathbf e_i)-V(\mathbf x)=-\int_0^hF_i(\mathbf x+s\mathbf e_i)\,ds$, so $\partial_iV=-F_i$.
The domain qualification is necessary and is implicit in the potential request. Curl-free alone on an arbitrary domain does not suffice: $\mathbf F=(-y,x,0)/(x^2+y^2)$ on the complement of the $z$-axis has zero <curl>, but its circulation around a unit circle is $2\pi$. It has no single-valued global potential. The intended construction is valid on all of $\mathbb R^3$ for a smooth curl-free field, and on any other domain satisfying the condition above.
For the given spherical-coordinate field, the angular components have $rF_\theta=\cos\theta\cos\phi$ and $rF_\phi=m\sin\phi$, independent of $r$, so the $\theta$ and $\phi$ <curl> components vanish. The radial component is
$$
(\nabla\times\mathbf F)_r
=\frac1{r\sin\theta}\left[\partial_\theta(\sin\theta F_\phi)-\partial_\phi F_\theta\right]
=\frac{(m+1)\cos\theta\sin\phi}{r^2\sin\theta}.
$$
It vanishes identically exactly when \b[$m=-1$]. In that case integrate $F_\theta=-r^{-1}V_\theta$ to obtain $V=-\sin\theta\cos\phi+C(\phi)$. The azimuthal equation $F_\phi=-(r\sin\theta)^{-1}V_\phi$ forces $C'\!=0$, and $F_r=0$ removes any radial dependence. At the given north-pole point the first term is zero, so normalization fixes the constant to zero:
$$
\boxed{V=-\sin\theta\cos\phi=-\frac{x}{\sqrt{x^2+y^2+z^2}}.}
$$
The Cartesian expression shows that this potential and the field $\mathbf F=\nabla(x/r)$ are smooth for $r>0$, including the spherical-coordinate poles. The origin is excluded because the specified field is singular there; that exclusion does not obstruct the displayed global potential.
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