Suppose we have a sphere defined as $x^2 + y^2 + z^2 = 9$. With radius of 3, we know the Gaussian Curvature should be $\frac{1}{9}$ at every point on the surface.
Since this defines a level set, the gradient gives the normals. So, the normals should be
$\begin{bmatrix} 2x \\\\ 2y \\\\ 2z \end{bmatrix}$
We can then normalize these normals, giving
$\begin{bmatrix} \frac{2x}{\sqrt{4x^2+4y^2+4z^2}} \\\\ \frac{2y}{\sqrt{4x^2+4y^2+4z^2}} \\\\ \frac{2z}{\sqrt{4x^2+4y^2+4z^2}} \\\\ \end{bmatrix}$
By my understanding, this gives us the surface normals on the unit sphere (the Gauss Map).
According to Wolfram MathWorld's article on Gauss Map, if we take the partial derivatives of the Gauss Map function, and then take the determinant of that matrix, we should get the curvature. The matrix in question is
$\begin{bmatrix} \frac{y^2+z^2}{{(x^2+y^2+z^2)}^{\frac{3}{2}}} & \frac{-xy}{{(x^2+y^2+z^2)}^{\frac{3}{2}}} & \frac{-xz}{{(x^2+y^2+z^2)}^{\frac{3}{2}}} \\\\ \frac{-xy}{{(x^2+y^2+z^2)}^{\frac{3}{2}}} & \frac{x^2+z^2}{{(x^2+y^2+z^2)}^{\frac{3}{2}}} & \frac{-yz}{{(x^2+y^2+z^2)}^{\frac{3}{2}}} \\\\ \frac{-xz}{{(x^2+y^2+z^2)}^{\frac{3}{2}}} & \frac{-yz}{{(x^2+y^2+z^2)}^{\frac{3}{2}}} & \frac{x^2+y^2}{{(x^2+y^2+z^2)}^{\frac{3}{2}}} \end{bmatrix}$
However, if we pick a point on the sphere, (0,3,0), then the resulting determinant is zero when we plug in that point. Where am I going wrong? The result should be $\frac{1}{9}$.