Firstly, rescale the variables as
$$
x\rightarrow ax \quad t\rightarrow bt \quad y\rightarrow cy
$$
and choose
$$
a=\frac{2h_0}{3}\sqrt{\frac{h_0}{g}}\quad b=h_0c\quad \frac{h_0^3c^2}{9}=1,
$$
the equation takes the form
$$
\frac{\partial x}{\partial t}+\lambda x\frac{\partial x}{\partial y}+\frac{\partial x}{\partial y}+\frac{\partial^3 x}{\partial y^3}=0
$$
being now $\lambda$ an ordering parameter, introduced by convenience, taken to be 1 at the end of the computation.
Then, you can consider two different situations: $\lambda\rightarrow 0$ and $\lambda\rightarrow\infty$. For the former case, take the series
$$
x=x_0+\lambda x_1+O(\lambda^2).
$$
I limit the computation at the first order just to explain the technique. Put this into the equation and you will obtain the set of equations
$$
\frac{\partial x_0}{\partial t}+\frac{\partial x_0}{\partial y}+\frac{\partial^3 x_0}{\partial y^3}=0,
$$
$$
\frac{\partial x_1}{\partial t}+\frac{\partial x_1}{\partial y}+\frac{\partial^3 x_1}{\partial y^3}=-x_0\frac{\partial x_1}{\partial y}-x_1\frac{\partial x_0}{\partial y},
$$
$$
\vdots.
$$
Now, you can realize that the equation for $x_0$ is linear and can be solved exactly, e.g. with a Fourier series, given the proper boundary and initial conditions. The procedure can be iterated at whatever order you like, just consider that going at higher orders implies more involved computations.
Finally, the opposite limit can be evaluated by rescaling the time variable as $t\rightarrow \lambda t$. Then, take the series
$$
x=x_0+\frac{1}{\lambda}x_1+O\left(\frac{1}{\lambda^2}\right).
$$
This kind of approach is known in fluid dynamics as a boundary layer problem. Then you get the set of equations
$$
\frac{\partial x_0}{\partial t}+x_0\frac{\partial x_0}{\partial y}=0,
$$
$$
\frac{\partial x_1}{\partial t}+x_0\frac{\partial x_1}{\partial y}+x_1\frac{\partial x_0}{\partial y}=-\frac{\partial x_0}{\partial y}-\frac{\partial^3x_0}{\partial y^3},
$$
$$
\vdots.
$$
I do not enter too much into these equations as in boundary layer problems there is a further complication arising by the boundary conditions. I just notice that the leading order equation can be solved by the characteristic method.
This should give you some starting point to work with.