Answer
$$ (a) \text{Degree of homogeneity: } n=0$$
$$ (b)\ \ x f_{x}(x, y)+y f_{y}(x, y)
=0=(0) f(x, y) $$
Work Step by Step
Given$$f(x, y)=e^{\frac{x}{y}}$$
\begin{array}{l}{\text { (a) since } } \\ { \qquad f(t x, t y)=e^{\frac{tx}{ty}}==e^{\frac{x}{y}}=t^{0} f(x, y) } \\ {\text { So, Degree of homogeneity: } }n=0
\end{array}
Since $$f_x(x,y)=\frac{\partial f(x,y)}{\partial y}= \frac{1}{y}e^{\frac{x}{y}}$$
$$f_y(x,y)=\frac{\partial f(x,y)}{\partial y}=-\frac{x}{y^2}e^{\frac{x}{y}}$$
So, we get
\begin{align} \text { (b) } x f_{x}(x, y)+y f_{y}(x, y)&=x\left(\frac{1}{y}e^{\frac{x}{y}}\right)+y\left(-\frac{x}{y^2}e^{\frac{x}{y}}\right)\\
& =\frac{x}{y}e^{\frac{x}{y}}-\frac{x}{y}e^{\frac{x}{y}}\\
&=0 \\
&=(0)f(x, y) \end{align}
