Answer
$$ (a) \text{Degree of homogeneity: } n=1$$
$$ (b)\ \ x f_{x}(x, y)+y f_{y}(x, y)
=1 f(x, y) $$
Work Step by Step
Given$$f(x, y)=\frac{x y}{\sqrt{x^{2}+y^{2}}}$$
it can be written $$\Rightarrow f(x, y)= x y (x^{2}+y^{2})^{-\frac{1}{2}} $$
\begin{array}{l}{\text { (a)- since } } \\ {\qquad f(t x, t y)=\frac{(t x)(t y)}{\sqrt{(t x)^{2}+(t y)^{2}}}= \frac{t^2x y}{t\sqrt{x^{2}+y^{2}}}=t\left(\frac{x y}{\sqrt{x^{2}+y^{2}}}\right)=t^1 f(x, y)} \\ {\text { So, degree of homogeneity: } }n=1
\end{array}
Since $$f_x(x,y)=\frac{\partial f(x,y)}{\partial y}=y (x^{2}+y^{2})^{-\frac{1}{2}} -\frac{1}{2}x y (2x) (x^{2}+y^{2})^{-\frac{3}{2}}\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =(x^{2}+y^{2})^{-\frac{3}{2}}\left( y (x^{2}+y^{2}) -x^2 y \right)=y^3(x^{2}+y^{2})^{-\frac{3}{2}} $$
$$f_y(x,y)=\frac{\partial f(x,y)}{\partial y}=x (x^{2}+y^{2})^{-\frac{1}{2}} -\frac{1}{2}x y (2y) (x^{2}+y^{2})^{-\frac{3}{2}}
\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =(x^{2}+y^{2})^{-\frac{3}{2}}\left( x (x^{2}+y^{2}) -x y^2 \right)=x^3(x^{2}+y^{2})^{-\frac{3}{2}} $$
\begin{array}{l} {\text { (b) } x f_{x}(x, y)+y f_{y}(x, y)=x\left(\frac{y^{3}}{\left(x^{2}+y^{2}\right)^{3 / 2}}\right)+y\left(\frac{x^{3}}{\left(x^{2}+y^{2}\right)^{3 / 2}}\right)\\
=xy\left(\frac{x^{2}+y^2}{\left(x^{2}+y^{2}\right)^{3 / 2}}\right)=\frac{x y}{\sqrt{x^{2}+y^{2}}}
=1 f(x, y)}\end{array}
