Answer
Using the Chain Rule we obtain Eq. (11):
$x\frac{{\partial f}}{{\partial x}} + y\frac{{\partial f}}{{\partial y}} + z\frac{{\partial f}}{{\partial z}} = nf$
Work Step by Step
Let $F\left( t \right) = f\left( {tx,ty,tz} \right)$. Since $f\left( {x,y,z} \right)$ is homogeneous of degree $n$, we have
(1) ${\ \ \ }$ $F\left( t \right) = f\left( {tx,ty,tz} \right) = {t^n}f\left( {x,y,z} \right)$
Write
$u\left( {x,y,z} \right) = tx$, ${\ \ }$ $v\left( {x,y,z} \right) = ty$, ${\ \ }$ $w\left( {x,y,z} \right) = tz$
We calculate $F'\left( t \right)$ using the Chain Rule:
$\frac{{dF}}{{dt}} = F'\left( t \right) = \frac{{\partial f}}{{\partial u}}\frac{{\partial u}}{{\partial t}} + \frac{{\partial f}}{{\partial v}}\frac{{\partial v}}{{\partial t}} + \frac{{\partial f}}{{\partial w}}\frac{{\partial w}}{{\partial t}}$
$F'\left( t \right) = x\frac{{\partial f}}{{\partial u}} + y\frac{{\partial f}}{{\partial v}} + z\frac{{\partial f}}{{\partial w}}$
However, from the right-hand side of equation (1), we also have
$F'\left( t \right) = n{t^{n - 1}}f\left( {x,y,z} \right)$
Therefore
$F'\left( t \right) = x\frac{{\partial f}}{{\partial u}} + y\frac{{\partial f}}{{\partial v}} + z\frac{{\partial f}}{{\partial w}} = n{t^{n - 1}}f\left( {x,y,z} \right)$
For $t=1$, we have $u=x$, $v=y$, $w=z$. So,
$F'\left( 1 \right) = x\frac{{\partial f}}{{\partial x}} + y\frac{{\partial f}}{{\partial y}} + z\frac{{\partial f}}{{\partial z}} = nf$
Hence,
$x\frac{{\partial f}}{{\partial x}} + y\frac{{\partial f}}{{\partial y}} + z\frac{{\partial f}}{{\partial z}} = nf$
