For a functions $f(x)$, $g(y)$ and $h(z)$, the chain rule yields

$\frac{\partial f(x)}{\partial \phi} = ( \frac{\partial f}{\partial x} ) ( \frac{\partial x}{\partial \phi} )$.

$\frac{\partial g(y)}{\partial \phi} = (\frac{\partial g}{\partial y}) (\frac{\partial y}{\partial \phi} )$.

$\frac{\partial h(z)}{\partial \phi} = (\frac{\partial h}{\partial z})( \frac{\partial z}{\partial \phi} )$.

Suppose $F(x,y,z) \equiv f(x) g(y) h(z)$.

This is a significant assumption.

Then, using the product rule,

$\frac{\partial F(x,y,z)}{\partial \phi} = \frac{\partial (f(x) g(y) h(z))}{\partial \phi} = g(y) h(z) \frac{\partial f(x)}{\partial \phi} + f(x) \frac{\partial [g(y) h(z)]}{\partial \phi}$.

Evaluate $\frac{\partial [g(y) h(z)]}{\partial \phi}$ using the product rule again:

$\frac{\partial [g(y) h(z)]}{\partial \phi} = h(z) \frac{\partial g(y) }{\partial \phi} + g(y) \frac{\partial h(z)}{\partial \phi}$.

Substituting,

$\frac{\partial F(x,y,z)}{\partial \phi} = g(y) h(z) \frac{\partial f(x)}{\partial \phi} + f(x) \bigg( h(z) \frac{\partial g(y) }{\partial \phi} + g(y) \frac{\partial h(z)}{\partial \phi} \bigg)$.

Distribute.

$\frac{\partial F(x,y,z)}{\partial \phi} = g(y) h(z) \frac{\partial f(x)}{\partial \phi} + f(x) h(z) \frac{\partial g(y) }{\partial \phi} + f(x) g(y) \frac{\partial h(z)}{\partial \phi}$.

Substitute:

$\frac{\partial F(x,y,z)}{\partial \phi} = g(y) h(z) ( \frac{\partial f}{\partial x} ) ( \frac{\partial x}{\partial \phi} ) + f(x) h(z) (\frac{\partial g}{\partial y}) (\frac{\partial y}{\partial \phi} ) + f(x) g(y) (\frac{\partial h}{\partial z})( \frac{\partial z}{\partial \phi} )$.

Rearranging:

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

Since multiplication is commutative,

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

Rewriting in a slightly different form,

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

Removing parentheses so that the rightmost partial derivatives do not look like isolated factors,

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

Rewriting with a distributive property:

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

Assuming the only function to the right of the differential operators is $F(x,y,z) \equiv f(x) g(y) h(z)$, one can write:

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

References

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