First 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

$\frac{\partial }{\partial \phi} = ( \frac{\partial x}{\partial \phi} ) \frac{\partial }{\partial x} + ( \frac{\partial y}{\partial \phi} ) \frac{\partial }{\partial y} + ( \frac{\partial z}{\partial \phi} ) \frac{\partial }{\partial z}$.

### References

[1] http://www.thphys.nuim.ie/Notes/MP469/Laplace.pdf