= Solution
For $t>0$, differentiate the printed ansatz:
$$
h=a t^{-1/3}-\frac{ax^2}{t}-bt,\qquad
h_t=-\frac a3t^{-4/3}+\frac{ax^2}{t^2}-b,
$$
$$
h_x=-\frac{2ax}{t},\qquad
(hh_x)_x=6a^2\frac{x^2}{t^2}-2a^2t^{-4/3}+2ab.
$$
Matching the independent coefficients in $h_t=(hh_x)_x-1$ gives
$$
a=6a^2,\qquad \frac a3=2a^2,\qquad b(1+2a)=1.
$$
There are two parameter pairs, including the degenerate spatially constant case:
$$
\boxed{(a,b)=\left(\frac16,\frac34\right)\quad\text{or}\quad(a,b)=(0,1)}.
$$
Their corresponding solutions are
$$
\boxed{h(x,t)=\frac1{6t^{1/3}}-\frac{x^2}{6t}-\frac34t}
\qquad\text{and}\qquad
\boxed{h(x,t)=-t}.
$$
The request for “the general form” cannot mean the complete solution set of this nonlinear <partial differential equation>: substitution of one ansatz cannot provide arbitrary initial data, and the <superposition principle> does not apply.
A useful wider family follows by allowing an independent time-dependent constant term and quadratic coefficient, $h=A(t)+B(t)x^2$. Coefficient comparison gives $B'=6B^2$ and $A'=2AB-1$. For $B\ne0$, write $\tau=t-t_0>0$ to obtain $B=-1/(6\tau)$ and $A=C\tau^{-1/3}-3\tau/4$. Spatial translation, also a symmetry of the equation, therefore gives
$$
\boxed{h(x,t)=C\tau^{-1/3}-\frac{(x-x_0)^2}{6\tau}-\frac34\tau,
\qquad\tau=t-t_0>0}.
$$
Here $C,x_0,t_0$ are arbitrary constants; the displayed nontrivial ansatz corresponds to $C=1/6$ and $x_0=t_0=0$. There are also affine solutions $h=px+(p^2-1)t+q$, including the spatially constant branch. These are valid solution families, rather than a claim of a full general solution.
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