Wow! My first Function Problem

Assuming $f(x)$ is differentiable over $x,y \in \mathbb{R}$ , let us differentiate this functional equation with respect to x and y:

$\frac{1}{3} \cdot f'(\frac{x+y}{3}) = \frac{f'(x)}{2} = \frac{f'(y)}{2} \Rightarrow f'(x) = f'(y) = A \Rightarrow f(x) = Ax + B$ (i).

Substituting (i) back into the original functional equation results in:

$A(\frac{x+y}{3}) + B = \frac{1}{2} \cdot [Ax + B + Ay + B] \Rightarrow (\frac{A}{3})(x+y) = (\frac{A}{2})(x+y) \Rightarrow A = 0$

Hence, our required function must be constant: $f(x) = B, B \in \mathbb{R}.$ The final integration produces $\int f(2x^{9}) dx = \int B dx = Bx$ with $L = M = N = 1$ and $\lfloor{1000M + 29N + 8L}\rfloor$ = $\boxed{1037}.$