Euler theorem says,
If $$u=f(x,y)\text{ ,homogeneous}$$ Then,$$x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=nu$$ Where $$n\to \text{degree of function}$$ Question
If $$u=u_1+u_2+u_3$$ then$$x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=(x\frac{\partial u_1}{\partial x}+y\frac{\partial u_1}{\partial y})+(x\frac{\partial u_2}{\partial x}+y\frac{\partial u_2}{\partial y})+(x\frac{\partial u_3}{\partial x}+y\frac{\partial u_3}{\partial y})$$ $$nu=n_1u_1+n_2u_2+n_3u_3$$ is it possible to apply the theorem like this.
one way to see the problem with your assumed form is to note that you would need : $$ nu =n(u_1+u_2+u_3) \\ =n_1u_1+n_2u_2+n_3u_3 $$ so $$ \sum_{j=1}^3(n-n_j)u_j = 0 $$