In the vicinity of a point on a 3-dimensional Riemannian manifold, let's consider a coordinate system $(x_1, x_2, x_3)$, where the metric tensor (or line element) is expressed in the following form.
\begin{equation*} ds^2=f(x_1,x_2,x_3)(d{x_1}^2+d{x_2}^2+d{x_3}^2). \end{equation*}
$A$ is a $3 \times 3$ rotation matrix. The coordinate transformation is defined as follows. \begin{equation*} \begin{pmatrix} y_1 \\ y_2 \\ y_3 \\ \end{pmatrix} =A \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ \end{pmatrix}. \end{equation*}
Then, for any $A$, the component of this metric tensor is preserved. \begin{equation*} ds^2=f(x_1,x_2,x_3)(d{y_1}^2+d{y_2}^2+d{y_3}^2). \end{equation*} Therefore, when the manifold is spherically symmetric, \begin{equation*} f(y_1,y_2,y_3)=f(x_1,x_2,x_3). \end{equation*} In this case, just like in Euclidean space, I think that $f(x, y, z)$ becomes a function of $x^2 + y^2 + z^2$. Can you prove it?