Consider the usual spherical coordinates:
(x,y,z)=(r \sin\theta \cos\phi, r \sin\theta \sin\phi, r \cos\theta )

find the superscript basis vectors: e^r, e^\theta, e^\phi
find the subscripted basis vectors: e_r, e_\theta, e_\phi
calculate the metric tensor gij

superscripted eU= grad(u)
subscripted   eu= partial_u <x,y>

<x,y,z>=<r Sin[theta] Cos[phi], r Sin[theta] Sin[phi], r Cos[theta]>

er=< Sin[theta] Cos[phi],  Sin[theta] Sin[phi],  Cos[theta]> = usual hat{r}
etheta=<r Cos[theta] Cos[phi], r Cos[theta] Sin[phi], - r Sin[theta]> = r usual hat{theta}
ephi=< - Sin[theta] Sin[phi],  Sin[theta] Cos[phi],  0> = r Sin[theta] usual hat{phi}

(r, theta, phi)=<Sqrt[x^2+y^2+z^2],ArcTan[Sqrt[x^2+y^2]/z], ArcTan[y/x]>

eR = grad(r)=<x,y,z>/r = usual hat{r}
eTHETA= = grad(ArcTan[Sqrt[x^2+y^2]/z])

In[1]:= {D[ArcTan[Sqrt[x^2+y^2]/z],x],D[ArcTan[Sqrt[x^2+y^2]/z],y],D[ArcTan[Sqrt[x^2+y^2]/z],z]}    

                                                                                2    2
                       x                              y                   Sqrt[x  + y ]
Out[1]= {-----------------------------, -----------------------------, -(----------------)}
                             2    2                         2    2             2    2
               2    2       x  + y            2    2       x  + y             x  + y    2
         Sqrt[x  + y ] (1 + -------) z  Sqrt[x  + y ] (1 + -------) z    (1 + -------) z
                               2                              2                  2
In[2]:= % /.{x->r Sin[theta] Cos[phi], y->r Sin[theta] Sin[phi], z->r Cos[theta]}       
           
In[3]:= FullSimplify[%,{r>0, Sin[theta]>0}]                                                         

         Cos[phi] Cos[theta]  Cos[theta] Sin[phi]    Sin[theta]
Out[3]= {-------------------, -------------------, -(----------)}
                  r                    r                 r

this is (1/r)* usual hat{theta}

ePHI= = grad(ArcTan[y/x])


In[7]:= { D[ArcTan[y/x],x],D[ArcTan[y/x],y],D[ArcTan[y/x],z]}                                       

                y            1
Out[7]= {-(-----------), ----------, 0}
                    2            2
            2      y            y
           x  (1 + --)   x (1 + --)
                    2            2
                   x            x

In[8]:= % /.{x->r Sin[theta] Cos[phi], y->r Sin[theta] Sin[phi]}                                    

           Csc[theta] Sec[phi] Tan[phi]   Csc[theta] Sec[phi]
Out[8]= {-(----------------------------), -------------------, 0}
                               2                          2
                r (1 + Tan[phi] )          r (1 + Tan[phi] )

In[9]:= FullSimplify[%]                                                                             

           Csc[theta] Sin[phi]   Cos[phi] Csc[theta]
Out[9]= {-(-------------------), -------------------, 0}
                    r                     r

this is 1/( r sin(theta) ) * usual hat{phi}

Note for example that ePHI.ephi=1 & ePHI.er=0 all as required

gij is the dot product of the lowered e.  The lowered e are mutually orthogonal
so the result is diagonal.  Note that each is proportial to the corresponding
unit vectors hat{}, so the dot product of the hats is one and we are left with
the proportionality constant

er = usual hat{r}
etheta= r usual hat{theta}
ephi= r Sin[theta] usual hat{phi}

so gij = diagonal{ 1, r^2, r^2 Sin[theta]}

which produces the usual result for differential distances:
dr^2 + r^2((d theta)^2 + sin(theta)^2 (d phi)^2)