There are basically no interesting solutions to this equation beyond first and zeroth order operators, even if one only imposes the stated constraint for $n=2$.

First, we can depolarise the hypothesis
$$ D^u(f^2) = \alpha_2 D^u(f) f \quad (1)$$
by replacing $f$ with $f+g, f-g$ for arbitrary functions $f,g$ and subtracting (and then dividing by $4$) to obtain the more flexible Leibniz type identity
$$ D^u(fg) = \frac{\alpha_2}{2}( D^u(f) g + f D^u(g) ). \quad (2)$$

There are now three cases, depending on the value of $\alpha_2$:

- $\alpha_2 \neq 1,2$. Applying (2) with $f=g=1$ we then conclude that $D^u(1)=0$, and then applying (2) again with just $g=1$ we get $D^u(f)=0$. So we have the trivial solution $D^u=0$ in this case.
- $\alpha_2=2$. Then $D^u$ is a derivation and by induction we have $D^u(f^n) = n D^u(f) f^{n-1}$, just as with the ordinary derivative, so we just have $\alpha_n=n$ for all $n$ with no fractional behavior.
- $\alpha_2=1$. Applying (2) with $g=1$ we obtain (after a little bit of algebra) $D^u(f) = mf$ where $m := D^u(1)$. Thus $D^u$ is just a multiplier operator, which obeys $D^u(f^n) = D^u(f) f^{n-1}$, thus $\alpha_n=1$ for all $n$.

Thus there are no linear solutions to your equation other than the usual derivations (e.g., $D^u(f) = a(x) \frac{d}{dx} f$ for any smooth symbol $a$) and multiplier operators $D^u(f) = mf$, i.e., first order and zeroth order operators.

On the other hand, fractional derivatives $D^u$ tend to obey a "fractional chain rule"
$$ D^u( F(f) ) = D^u(f) F'(f) + E$$
for various smooth functions $F,f$, where the error $E$ obeys better estimates in various Sobolev spaces than the other two terms in this equation. In particular, for $F(t) = t^n$, we would have
$$ D^u(f^n) = n D^u(f) f^{n-1} + E$$
for a "good" error term $E$. For instance, taking $u=n=2$ with $D$ the usual derivative, we have
$$ D^2(f^2) = 2 D^2(f) f + E \quad (3)$$
with $E$ the "carré du champ" operator
$$ E := 2 (Df)^2.$$
Note that the error $E$ is controlled uniformly by the $C^1$ norm of $f$ but the other two terms in (3) are not. See my previous MathOverflow answer at https://mathoverflow.net/a/94039/766 for some references and further discussion.