For the sake of simplicity, let me pick $f(u_x)=\frac{1}{2}u_x^2$ and $u_0\in H^5$. We have that, in the new variable $f=u_x$, the equation reads
$$
f_t-\alpha f_{xx}-\beta f_{xxx}=ff_x
$$

We have that
$$
\frac{1}{2}\frac{d}{dt}\|f\|_{L^2}^2+\alpha\|f_x\|_{L^2}^2=0,
$$
so
$$
\|f(t)\|_{L^2}^2+2\alpha\int_0^t\|f_{x}\|_{L^2}^2ds\leq \|f(0)\|_{L^2}^2.
$$
If we test the equation against $-f_{xx}$, then
$$
\frac{1}{2}\frac{d}{dt}\|f_x\|_{L^2}^2+\alpha\|f_{xx}\|_{L^2}^2\leq \frac{1}{2\alpha}\|ff_x\|_{L^2}^2+\frac{\alpha}{2}\|f_{xx}\|_{L^2}^2.
$$
Using that
$$
\|ff_x\|_{L^2}^2\leq \|f\|_{L^4}^2\|f_x\|_{L^4}^2\leq C\|f\|_{H^{0.25}}^2\|f\|_{H^{1.25}}^2\leq C\|f\|_{L^2}^{3/2}\|f_{x}\|_{L^2}^{1/2}\|f_x\|_{L^2}^{3/2}\|f_{xx}\|_{L^2}^{1/2},
$$
thus,
$$
\|ff_x\|_{L^2}^2\leq C\|f_{x}\|_{L^2}^{4}+\frac{\alpha}{2}\|f_{xx}\|_{L^2}^{1/2}.
$$
Using Gronwall and the finiteness of $\int_0^t\|f_{x}\|_{L^2}^2ds$, we conclude. In this way, you can get a priori estimates for any Sobolev norm. After that you can just mollify your equation appropriately and pass to the limit in the approximate problems.