Here is the parametric description of the sphere.
\begin{equation*}
\sigma (\theta, \phi) = (R \sin \phi \cos \theta, R \sin
\phi \sin \theta, R \cos \phi)
\end{equation*}
I’ll calculate the partial derivative and take the cross product to calculate the normal.
\begin{align*}
\sigma_{\theta} \amp = \left(-R \sin \phi \sin \theta, R \sin
\phi \cos \theta, 0 \right) \\
\sigma_{\phi} \amp = \left( R \cos \phi \cos \theta, R \cos
\phi \sin \theta, -R \sin \phi \right) \\
\sigma_{\theta} \times \sigma_{\phi} \amp = \left( -R^2
\sin^2 \phi \cos \theta, -R^2 \sin^2 \phi \sin \theta,
-R^2 \sin \phi \cos \phi \right)\\
\amp = -R \sin \phi \left( R \sin \phi \cos \theta, R \sin
\phi \sin \theta, R \cos \phi) \right)
\end{align*}
I factored out the common term in the normal. After factoring out this, I see that the resulting normal is precisely same as the original position. So, the normal is a multiple of the position. This makes sense: for a sphere centred at the origin, the vector that points out of the sphere at point (as a local direction vector) has the same direction as the vector from the origin to that point.
The normal is \((0,0,0)\) when \(\phi = 0\text{,}\) which happens at the top and bottom poles of the sphere. These points are singularities of the parametrization. This was the same problem for spherical coordinates: the longitude is not propely defined at the north and south poles. The sphere itself is a surface without sharp corners, but the system of parametrization is the problem. It breaks down at the poles.