Answer
(a) $f\left( {x,y,z} \right)$ is homogeneous of degree 3.
(b) $f\left( {x,y,z} \right)$ is homogeneous of degree 1.
(c) $f\left( {x,y,z} \right)$ is homogeneous of degree 0.
(d) $f\left( {x,y,z} \right)$ is homogeneous of degree 4.
Work Step by Step
(a) We have $f\left( {x,y,z} \right) = {x^2}y + xyz$. So,
$f\left( {\lambda x,\lambda y,\lambda z} \right) = {\lambda ^3}{x^2}y + {\lambda ^3}xyz = {\lambda ^3}\left( {{x^2}y + xyz} \right)$
Since $f\left( {\lambda x,\lambda y,\lambda z} \right) = {\lambda ^n}f\left( {x,y,z} \right)$ for all $\lambda \in \mathbb{R}$, $f\left( {x,y,z} \right)$ is homogeneous of degree 3.
(b) We have $f\left( {x,y,z} \right) = 3x + 2y - 8z$. So,
$f\left( {\lambda x,\lambda y,\lambda z} \right) = 3\lambda x + 2\lambda y - 8\lambda z = \lambda \left( {3x + 2y - 8z} \right)$
Since $f\left( {\lambda x,\lambda y,\lambda z} \right) = \lambda f\left( {x,y,z} \right)$ for all $\lambda \in \mathbb{R}$, $f\left( {x,y,z} \right)$ is homogeneous of degree 1.
(c) We have $f\left( {x,y,z} \right) = \ln \left( {\frac{{xy}}{{{z^2}}}} \right)$. So,
$f\left( {\lambda x,\lambda y,\lambda z} \right) = \ln \left( {\frac{{{\lambda ^2}xy}}{{{\lambda ^2}{z^2}}}} \right) = \ln \left( {\frac{{xy}}{{{z^2}}}} \right)$
$f\left( {\lambda x,\lambda y,\lambda z} \right) = f\left( {x,y,z} \right)$
Since $f\left( {\lambda x,\lambda y,\lambda z} \right) = {\lambda ^0}f\left( {x,y,z} \right)$ for $\lambda \ne 0$, $f\left( {x,y,z} \right)$ is homogeneous of degree 0.
(d) We have $f\left( {x,y,z} \right) = {z^4}$. So,
$f\left( {\lambda x,\lambda y,\lambda z} \right) = {\lambda ^4}{z^4}$
Since $f\left( {\lambda x,\lambda y,\lambda z} \right) = {\lambda ^4}f\left( {x,y,z} \right)$ for all $\lambda \in \mathbb{R}$, $f\left( {x,y,z} \right)$ is homogeneous of degree 4.