Answer
(a). $f(x,y)=x^{2}y+2xy^{2}+5y^{3}$ is homogeneous of degree 3.
(b). $x\frac{\partial{f}}{\partial{x}}+y\frac{\partial{f}}{\partial{y}}=nf(x,y)$
Work Step by Step
(a). We know $f(x,y)=x^{2}y+2xy^{2}+5y^{3}$ is a polynomial, thus it has continuous second-order partial derivatives. Moreover,
$$f(tx,ty)=(tx)^{2}(ty)+2(tx)(ty)^{2}+5(ty)^{3}=t^{3}x^{2}y+2t^{3}xy^{2}+5t^{3}y^{3}=t^{3}(x^{2}y+2xy^{2}+5y^{3})=t^{3}f(x,y)$$
Therefore $f$ is homogeneous of degree 3.
(b). We have $f(tx,ty)=t^{n}f(x,y)$, by the Chain rule, differentiate both sides with respect to t. We obtain,
$$\frac{\partial{f(tx,ty)}}{\partial{t}}=\frac{\partial{[t^{n}f(x,y)]}}{\partial{t}}$$
$$\Rightarrow\frac{\partial{f(tx,ty)}}{\partial{(tx)}}\cdot\frac{\partial{(tx)}}{\partial{t}}+\frac{\partial{f(tx,ty)}}{\partial{(ty)}}\cdot\frac{\partial{(ty)}}{\partial{t}}=nt^{n-1}f(x,y)$$
$$\Rightarrow x\cdot\frac{\partial{f(tx,ty)}}{\partial{(tx)}}+y\cdot\frac{\partial{f(tx,ty)}}{\partial{(ty)}}=nt^{n-1}f(x,y)$$
Choose $t=1$, thus, we get
$$x\cdot\frac{\partial{f(x,y)}}{\partial{x}}+y\cdot\frac{\partial{f(x,y)}}{\partial{y}}=nf(x,y)$$
