Answer
$\frac{{\partial f}}{{\partial \phi }}{|_P} = 1 - 8\sqrt 2 $
Work Step by Step
We have $f\left( {x,y,z} \right)$, where $x = \rho \cos \theta \sin \phi $, $y = \rho \sin \theta \sin \phi $, $z = \rho \cos \phi $.
The partial derivatives of $x$, $y$ and $z$ are
$\frac{{\partial x}}{{\partial \rho }} = \cos \theta \sin \phi $, ${\ \ \ }$ $\frac{{\partial y}}{{\partial \rho }} = \sin \theta \sin \phi $, ${\ \ \ }$ $\frac{{\partial z}}{{\partial \rho }} = \cos \phi $
$\frac{{\partial x}}{{\partial \theta }} = - \rho \sin \theta \sin \phi $, ${\ \ \ }$ $\frac{{\partial y}}{{\partial \theta }} = \rho \cos \theta \sin \phi $, ${\ \ \ }$ $\frac{{\partial z}}{{\partial \theta }} = 0$
$\frac{{\partial x}}{{\partial \phi }} = \rho \cos \theta \cos \phi $, ${\ \ \ }$ $\frac{{\partial y}}{{\partial \phi }} = \rho \sin \theta \cos \phi $, ${\ \ \ }$ $\frac{{\partial z}}{{\partial \phi }} = - \rho \sin \phi $
Using the Chain Rule, Eq. (2) and Eq. (3) of Section 15.6, we have
$\frac{{\partial f}}{{\partial \rho }} = \frac{{\partial f}}{{\partial x}}\frac{{\partial x}}{{\partial \rho }} + \frac{{\partial f}}{{\partial y}}\frac{{\partial y}}{{\partial \rho }} + \frac{{\partial f}}{{\partial z}}\frac{{\partial z}}{{\partial \rho }}$
$\frac{{\partial f}}{{\partial \theta }} = \frac{{\partial f}}{{\partial x}}\frac{{\partial x}}{{\partial \theta }} + \frac{{\partial f}}{{\partial y}}\frac{{\partial y}}{{\partial \theta }} + \frac{{\partial f}}{{\partial z}}\frac{{\partial z}}{{\partial \theta }}$
$\frac{{\partial f}}{{\partial \phi }} = \frac{{\partial f}}{{\partial x}}\frac{{\partial x}}{{\partial \phi }} + \frac{{\partial f}}{{\partial y}}\frac{{\partial y}}{{\partial \phi }} + \frac{{\partial f}}{{\partial z}}\frac{{\partial z}}{{\partial \phi }}$
So,
$\frac{{\partial f}}{{\partial \rho }} = {f_x}\cos \theta \sin \phi + {f_y}\sin \theta \sin \phi + {f_z}\cos \phi $
$\frac{{\partial f}}{{\partial \theta }} = - {f_x}\rho \sin \theta \sin \phi + {f_y}\rho \cos \theta \sin \phi $
$\frac{{\partial f}}{{\partial \phi }} = {f_x}\rho \cos \theta \cos \phi + {f_y}\rho \sin \theta \cos \phi - {f_z}\rho \sin \phi $
At $P = \left( {\rho ,\theta ,\phi } \right) = \left( {2,\frac{\pi }{4},\frac{\pi }{4}} \right)$, we are given:
${f_x}\left( P \right) = 4$, ${\ \ \ }$ ${f_y}\left( P \right) = - 3$, ${\ \ \ }$ ${f_z}\left( P \right) = 8$
So,
$\frac{{\partial f}}{{\partial \phi }}{|_P} = 4\cdot2\cdot\cos \frac{\pi }{4}\cos \frac{\pi }{4} + \left( { - 3} \right)\cdot2\cdot\sin \frac{\pi }{4}\cos \frac{\pi }{4} - 8\cdot2\cdot\sin \frac{\pi }{4}$
$\frac{{\partial f}}{{\partial \phi }}{|_P} = 1 - 8\sqrt 2 $
