Answer
We use the limit definition of Section 15.3 to prove that ${f_x}\left( {x,y,z} \right)$ is homogeneous of degree $n-1$.
Work Step by Step
Let $f\left( {x,y,z} \right)$ be homogeneous of degree $n$.
Using the limit definition of Section 15.3, we have the partial derivative
${f_x}\left( {x,y,z} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h,y,z} \right) - f\left( {x,y,z} \right)}}{h}$
Next, we evaluate the limit:
${f_x}\left( {\lambda x,\lambda y,\lambda z} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {\lambda x + \lambda h,\lambda y,\lambda z} \right) - f\left( {\lambda x,\lambda y,\lambda z} \right)}}{{\lambda h}}$
Since $f\left( {x,y,z} \right)$ is homogeneous of degree $n$, we have
$f\left( {\lambda x,\lambda y,\lambda z} \right) = {\lambda ^n}f\left( {x,y,z} \right)$
$f\left( {\lambda x + \lambda h,\lambda y,\lambda z} \right) = f\left( {\lambda \left( {x + h} \right),\lambda y,\lambda z} \right) = {\lambda ^n}f\left( {x + h,y,z} \right)$
Thus, the partial derivative becomes
${f_x}\left( {\lambda x,\lambda y,\lambda z} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{{\lambda ^n}f\left( {x + h,y,z} \right) - {\lambda ^n}f\left( {x + h,y,z} \right)}}{{\lambda h}}$
$ = {\lambda ^{n - 1}}\mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h,y,z} \right) - f\left( {x,y,z} \right)}}{h}$
$ = {\lambda ^{n - 1}}{f_x}\left( {x,y,z} \right)$
Hence, by definition ${f_x}\left( {x,y,z} \right)$ is homogeneous of degree $n-1$.
