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 $$f(x, y)=\frac{x^{2}}{\sqrt{x^{2}+y^{2}}}$$
it can be written as
$$f(x, y)= x^{2} (x^{2}+y^{2} )^{-\frac{1}{2}}$$
$(a)$ Since
$f(t x, t y)=\frac{(t x)^{2}}{\sqrt{(t x)^{2}+(t y)^{2}}}=\frac{t^2 x^{2}}{t\sqrt{x^{2}+y^{2}}}=t\left(\frac{x^{2}}{\sqrt{x^{2}+y^{2}}}\right)=t f(x, y)$
$$ \text{So, Degree of homogeneity:} \ \ n =1$$
Since
\begin{align}
f_x(x,y)=\frac{\partial f(x,y)}{\partial x}&= 2x (x^{2}+y^{2} )^{-\frac{1}{2}}-\frac{1}{2}x^{2} (2x) (x^{2}+y^{2} )^{-\frac{3}{2}}\\&
= (x^{2}+y^{2} )^{-\frac{3}{2}}(2x (x^{2}+y^{2} )-x^{.3} )\\
&=(x^{2}+y^{2} )^{-\frac{3}{2}}(x^{3}+2x y^{2} )
\end{align}
\begin{align}
f_y(x,y)=\frac{\partial f(x,y)}{\partial y}&= 0-\frac{1}{2}x^{2} (2y) (x^{2}+y^{2} )^{-\frac{3}{2}}\\&
= -x^2 y (x^{2}+y^{2} )^{-\frac{3}{2}}
\end{align}
\begin{align}
&x f_{x}(x, y)+y f_{y}(x, y)=x(x^{2}+y^{2} )^{-\frac{3}{2}}(x^{3}+2x y^{2} )-y(x^2 y)(x^{2}+y^{2} )^{-\frac{3}{2}}\\
&=(x^{2}+y^{2} )^{-\frac{3}{2}}(x^{4}+2x^2 y^{2} -x^2 y^2)\\
&=(x^{2}+y^{2} )^{-\frac{3}{2}}(x^{4}+ x^2 y^{2} )\\
&=x^2 (x^{2}+y^{2} )^{-\frac{3}{2}}(x^{2}+ y^{2} )\\
&=x^2 (x^{2}+y^{2} )^{-\frac{1}{2}}\\
&=1 f(x, y)\end{align}
