If we define the free surface by:
$F(x,y,t)=y-h(x,t)=0$
Then for this to be a material surface
$\frac{DF}{Dt}=0$ on $y=h(x,t)$
However on $y=h(x,t)$, $F=0$, so doesn't this just imply $\frac{DF}{Dt}=0$ anyway? How does the fact that we are dealing with a material surface change anything here?