Consider the time independent case.
Then $F \equiv C$ is a solution to the equation, since $F_x = F_{xxx} = 0$.
A time-independent transformation gives $ C = f^2 + f_x \implies f_x = C - f^2 \tag{1}$ So that $ f_{xx} = -2 f f_x = -2C f + 2f^3 $ and $ f_{xxx} = -2C f_x + 6f^2 f_x = -2C^2 + 8C f^2 - 6f^4 $
On the other hand we have $ 6f^2 f_x = 6f^2 C - 6f^4 $ so $ f_t - 6f^2 f_x + f_{xxx} = -2C f_x = -2C^2 + 2C f^2 $ which given the ODE (1) can only vanish identically if either
- $C = 0$, or
- $f^2 \equiv C$.
So in the case $C < 0$, since we cannot have it being a perfect square ($f$ being real valued) we conclude that $f$ cannot solve the given equation.
To get a real concrete function: let $C = -1$, by direct integration we see that $f = \cot (x)$ satisfies $f_x = -1 - f^2$ and so corresponds to the solution $F \equiv -1$ of the KdV equation, but $ f_t - 6 f^2 f_x + f_{xxx} = -2 \csc(x) \neq 0 $