Christina DanielFirst focus on the Cartesian coordinate $x$, which depends on the spherical coordinates $r,\theta,$ and $\phi$.
For a function $f(x)$, the chain rule yields
$ \frac{\partial f(x)}{\partial r} = ( \frac{\partial f}{\partial x} ) ( \frac{\partial x}{\partial r} )$.
Similarly for functions $g(y)$ and $h(z)$:
$ \frac{\partial g(y)}{\partial r} = (\frac{\partial g}{\partial y}) (\frac{\partial y}{\partial r} )$.
$ \frac{\partial h(z)}{\partial r} = (\frac{\partial h}{\partial z})( \frac{\partial z}{\partial r} )$.
Suppose $F(x,y,z) \equiv f(x) g(y) h(z)$. This is a significant assumption.
Then, using the product rule twice,
$ \frac{\partial F(x)}{\partial r} = \frac{\partial}{\partial r} [f(x) g(y) h(z)] = g(y) h(z) \frac{\partial f(x)}{\partial r} + f(x) [ h(z) \frac{\partial g(y) }{\partial r} + g(y) \frac{\partial h(z) }{\partial r} ] $.
Simplifying,
$ \frac{\partial F(x,y,z) }{\partial r} = g(y) h(z) \frac{\partial f(x)}{\partial r} + f(x) h(z) \frac{\partial g(y) }{\partial r} + f(x) g(y) \frac{\partial h(z) }{\partial r} $.
Substituting,
$ \frac{\partial F(x,y,z) }{\partial r} = g(y) h(z) \frac{\partial f}{\partial x} \frac{\partial x}{\partial r} + f(x) h(z) \frac{\partial g}{\partial y} \frac{\partial y}{\partial r} + f(x) g(y) \frac{\partial h}{\partial z} \frac{\partial z}{\partial r} $.
Rearranging,
$ \frac{\partial F(x,y,z)}{\partial r} = \frac{\partial x}{\partial r} (\frac{\partial f}{\partial x}) ( g(y) h(z) ) + \frac{\partial y}{\partial r} ( \frac{\partial g}{\partial y} ) (f(x) h(z) ) + \frac{\partial z}{\partial r} ( \frac{\partial h}{\partial z}) ( f(x) g(y) )$.
Each partial derivative only depends on one variable, so it is possible to write:
$ \frac{\partial F(x,y,z) }{\partial r} = \frac{\partial x}{\partial r} (\frac{\partial f(x) g(y) h(z)}{\partial x}) + \frac{\partial y}{\partial r} ( \frac{\partial f(x) h(z) g(y)}{\partial y} ) + \frac{\partial z}{\partial r} ( \frac{\partial f(x) g(y) h(z) }{\partial z}) $.
Since $F(x,y,z) \equiv f(x) g(y) h(z)$,
$ \frac{\partial F(x,y,z) }{\partial r} = ( \frac{\partial x}{\partial r} ) \frac{\partial F(x,y,z) }{\partial x} + ( \frac{\partial y}{\partial r} ) \frac{\partial F(x,y,z) }{\partial y} + ( \frac{\partial z}{\partial r} )\frac{\partial F(x,y,z) }{\partial z} $.
Rewriting,
$ \frac{\partial F(x,y,z)}{\partial r} = ( \frac{\partial x}{\partial r} ) \frac{\partial }{\partial x} F(x,y,z) + ( \frac{\partial y}{\partial r} ) \frac{\partial }{\partial y} F(x,y,z) + ( \frac{\partial z}{\partial r} ) \frac{\partial }{\partial z} F(x,y,z) $.
Rewriting again,
$ \frac{\partial }{\partial r} F(x,y,z) = \bigg[ ( \frac{\partial x}{\partial r} ) \frac{\partial }{\partial x} + ( \frac{\partial y}{\partial r} ) \frac{\partial }{\partial y} + ( \frac{\partial z}{\partial r} ) \frac{\partial }{\partial z} \bigg] F(x,y,z) $.
Removing the $F(x,y,z)$ from the right,
$ \frac{\partial }{\partial r} = ( \frac{\partial x}{\partial r} ) \frac{\partial }{\partial x} + ( \frac{\partial y}{\partial r} ) \frac{\partial }{\partial y} + ( \frac{\partial z}{\partial r} ) \frac{\partial }{\partial z} $.
The same procedure can be done for $\frac{\partial }{\partial \theta} $ and $\frac{\partial }{\partial \phi} $.
Alternatively, consider the result:
$ \frac{\partial }{\partial r} = ( \frac{\partial x}{\partial r} ) \frac{\partial }{\partial x} + ( \frac{\partial y}{\partial r} ) \frac{\partial }{\partial y} + ( \frac{\partial z}{\partial r} ) \frac{\partial }{\partial z} $.
By analogy, it is possible to write, by replacing $r$ with the spherical coordinate of interest:
$ \frac{\partial }{\partial \theta} = ( \frac{\partial x}{\partial \theta} ) \frac{\partial }{\partial x} + ( \frac{\partial y}{\partial \theta} ) \frac{\partial }{\partial y} + ( \frac{\partial z}{\partial \theta} ) \frac{\partial }{\partial z} $.
and
$ \frac{\partial }{\partial \phi} = ( \frac{\partial x}{\partial \phi} ) \frac{\partial }{\partial x} + ( \frac{\partial y}{\partial \phi} ) \frac{\partial }{\partial y} + ( \frac{\partial z}{\partial \phi} ) \frac{\partial }{\partial z} $.