Operation on a vector with a special property

Question

I have a vector $\vec u$ with the following property: $$\vec u (tx, ty, tz) = t^n \vec u (tx, ty, tz)$$ Now I have to prove that: $$(\vec r.\vec \nabla)\vec u = n.\vec u$$ How can I do this? I tried to do it in the following way: $$(\vec r.\vec \nabla)\vec u = (x.\partial /\partial x + y.\partial/\partial y + z.\partial/\partial z) \vec u(tx, ty, tz)$$ $$ = (x.\partial/\partial x + y.\partial/\partial y + z.\partial/\partial z)[t^n \vec u(x, y, z)]$$ $$ = t^n(x\partial\vec u(x)/\partial x + y.\partial \vec u(y)/\partial y + z.\partial \vec u(z)/\partial z)$$ since under partial differentiation $\vec u(x, y, z)$ will get separated into $\vec u(x), \vec u(y), \vec u(z)$ for $\partial/\partial x, \partial/\partial y, \partial/\partial z$ respectively, and $t$ is a scalar. But I can't go any further. What is/are the mistake(s) in this approach, if any? Also, how can it be proven?