Answer:
Degree [tex]k=3[/tex]
Step-by-step explanation:
The function that you have is a homogeneous function of degree [tex]k=3[/tex]. In fact, if [tex]\lambda \in \mathbb{R}[/tex] and [tex](x,y) \in \mathbb{R}^{2}[/tex] we have that:
[tex]s(\lambda x, \lambda y)=(\lambda y)(\lambda x)^{2}+(\lambda x)(\lambda y)^{2}=\lambda^{3}x^{2}y+\lambda^{3}xy^{2}=\lambda^{3}s(x,y)[/tex]
To verify that the Euler Theorem holds we need to show that:
[tex]x\frac{\partial s}{\partial x}+y\frac{\partial s}{\partial y}=3 s(x,y)[/tex].
To do that let's compute the partial derivatives:
[tex]\frac{\partial s}{\partial x}=2xy+y^2[/tex]
[tex]\frac{\partial s}{\partial y}=x^2+2xy[/tex]
Then,
[tex]x\frac{\partial s}{\partial x}+y\frac{\partial s}{\partial y}=2x^{2}y+xy^{2}+yx^{2}+2xy^{2}=3xy^{2}+3x^{2}y=3(x^{2}y+xy^{2})=3s(x,y).[/tex]
