Answer
$$ (a) \text{Degree of homogeneity: } n=3$$
$$ (b)\ \ x f_{x}(x, y)+y f_{y}(x, y)
=3 f(x, y) $$
Work Step by Step
Given$$f(x, y)=x^{3}-3 x y^{2}+y^{3}$$
\begin{array}{l}{\text { (a) since } } \\ { \qquad f(t x, t y)=(t x)^{3}-3(t x)(t y)^{2}+(t y)^{3}=t^{3}\left(x^{3}-3 x y^{2}+y^{3}\right)=t^{3} f(x, y) } \\ {\text { So, degree of homogeneity: } }n=3
\end{array}
Since $$f_x(x,y)=\frac{\partial f(x,y)}{\partial y}= 3x^2-3y^2$$
$$f_y(x,y)=\frac{\partial f(x,y)}{\partial y}=-6xy+3y^2$$
So, we get
\begin{align} \text { (b) } x f_{x}(x, y)+y f_{y}(x, y)&=x\left(3x^2-3y^2\right)+y\left(-6xy+3y^2\right)\\
& = 3x^3-3 xy^2 -6xy^2+3y^3\\
&= 3x^3-9 xy^2 +3y^3 \\
& = 3(x^3-3 xy^2 + y^3)\\
&=3f(x, y) \end{align}
