Cartesian to Spherical coordinate system - Coordinate Transformation

In general orthogonal curvilinear coordinate system, general position in 3 dimension is given by, $$\vec{s} = \vec{s}(u_1, u_2, u_3)$$ and small displacement "ds" is can be written as, $$\vec{ds} = \sum_i^3 \frac{\partial{\vec{s}}}{\partial{u_i}} du_i$$ where $ \frac{\partial{\vec{s}}}{\partial{u_i}} = h_i \hat{e_i} $ and $\hat{e_i} $ is the unit vector along the 'i'th direction. Rewriting it in simple form, we have, $$\vec{ds} = \sum_i^3 h_i du_i \hat{e_i}$$
For cartesian coordinate system, $h_i = 1$
$\rightarrow$ 
$$\vec{ds} = dx \hat{e_x} + dy \hat{e_y} + dz \hat{e_z}$$
But there is no unique choice of coordinate system, we can also choose spherical coordinate system as, 
$$\vec{ds} = dr \hat{e_r} + r \,d\theta\hat{e_\theta} + r \,sin\theta \,d\phi \hat{e_\phi}$$ with corresponding scaling factors. 

To go from one coordinate system to another, we use the relations, $$x = x(r,\theta,\phi) \\~\\ y = y(r,\theta,\phi) \\~\\ z=z(r,\theta, \phi)$$  
The only relation we know from our conventional assumption is that, $$x = \,r \,sin\theta \,cos\phi \\~\\ y = \,r \,sin\theta \,sin\phi \\~\\ z = \,r \,cos\theta\,$$  
Using the chain rule, $$dx = \frac{\partial{x}}{\partial{r}}dr + \frac{\partial{x}}{\partial{\theta}} d\theta + \frac{\partial{x}}{\partial{\phi}}d\phi$$
$\rightarrow$ $$dx = \,sin\theta \,cos\phi \,dr + r \,cos\theta\, cos\phi \,d\theta - r \,sin\theta \,sin\phi \,d\phi$$
Similarly, $$dy = \, sin\theta \, sin\phi \, dr + r \, cos\theta \, sin\phi \, d\theta + \, r\, sin\theta\, cos\phi \, d\phi$$ and $$dz = cos\theta \, dr - r\, sin\theta \, d\theta$$
Applying it in our Cartesian equation for "ds", we get, $$ds =(\,sin\theta \,cos\phi \,dr + r \,cos\theta\, cos\phi \,d\theta - r \,sin\theta \,sin\phi \,d\phi )\hat{e_x} \\~\\+ (\, sin\theta \, sin\phi \, dr + r \, cos\theta \, sin\phi \, d\theta + \, r\, sin\theta\, cos\phi \, d\phi)\hat{e_y} \\~\\+ (cos\theta \, dr - r\, sin\theta \, d\theta) \hat{e_z}$$  

Thus we expressed the infinitesimal displacement in cartesian system using spherical measurements such $ r, \theta, \phi $. Now, from our definition of unit vector, $$\hat{e_r} = \frac{{\frac{\partial{\vec{s}}}{\partial{r}}}}{|\frac{\partial{\vec{s}}}{\partial{r}}|}$$ and $$\hat{e_\theta} = \frac{{\frac{\partial{\vec{s}}}{\partial{\theta}}}}{|\frac{\partial{\vec{s}}}{\partial{\theta}}|}$$ and $$\hat{e_\phi} = \frac{{\frac{\partial{\vec{s}}}{\partial{\phi}}}}{|\frac{\partial{\vec{s}}}{\partial{\phi}}|}$$  
Using the above, equation, we can write the unit vectors along $r,\,\theta,\,\phi $ directions as follows, $$\hat{e_r} = sin\theta\, cos\phi\, \hat{e_x} \,+ \,\sin\theta\,sin\phi\,\hat{e_y} \,+\, cos\theta\,\hat{e_z}$$ and $$\hat{e_\theta} = cos\theta\, cos\phi \,\hat{e_x} \,+\, \cos\theta\, sin\phi \,\hat{e_y} - sin\theta\,\hat{e_z}$$ and $$\hat{e_\phi} = \,-sin\phi \,\hat{e_x} \,+ \,cos\phi\,\hat{e_y}$$ From these three unit vectors we can solve for the other three unit vectors as following,
Solving the first two,
  $$sin\theta\,\hat{e_r} + \, cos\theta\,\hat{e_\theta} = \,cos\phi \hat{e_x} + \,sin\phi \,\hat{e_y}$$

Combining with the third we can solve for x and y, $$\hat{e_x} = sin\theta\,cos\phi\,\hat{e_r} + cos\theta\,cos\phi\,\hat{e_\theta} - \,sin\phi\,\hat{e_\phi}$$ and $$\hat{e_y} = \,sin\theta\,sin\phi \hat{e_r} +\, cos\theta\, sin\phi\, \hat{e_\theta} + \,cos\phi\, \hat{e_\phi}$$ and finally solving for z, $$\hat{e_z} = cos\theta\,\hat{e_r} - sin\theta\,\hat{e_\theta}$$

That is all we do need to derive for the coordinate transformation from cartesian to spherical.