So anyone can tell me specifically, what $(v \cdot \nabla)f(x, y ,t) $ is?
Original question is:
$\phi$ is velocity potential, $h(x, t)$ is the depth, flow velocity is $v = \nabla \phi$. Define $f(x, y, t) = h(x, t) -y $. How to get $$\frac{df}{dt} = \frac{\partial f}{\partial t} + (v\cdot \nabla)f ,$$ further, how to get $$\frac{\partial h}{\partial t} + \frac{\partial \phi}{\partial x} \frac{\partial h}{\partial x} - \frac{\partial \phi}{\partial y} = 0$$
Using the chain rule for $f(x,y,t)$, we have $$ \frac{d f}{d t} = \frac{\partial x}{\partial t} \frac{\partial f}{\partial x} + \frac{\partial y}{\partial t} \frac{\partial f}{\partial y} + \frac{\partial f}{\partial t} = v\cdot \nabla f + \frac{\partial f}{\partial t} $$ where $v = \partial_t (x, y)^\top$ and $\nabla f = (\partial_x f, \partial_y f)^\top$. For the particular case $f(x,y,t) = h(x,t) - y$, the computation of the gradient gives $\nabla f = (\partial_x h, -1)^\top$, so that $$ \frac{d f}{d t} = \frac{\partial \phi}{\partial x} \frac{\partial h}{\partial x} - \frac{\partial \phi}{\partial y} + \frac{\partial h}{\partial t} \, , $$ where $v = (\partial_x\phi, \partial_y\phi)^\top = \nabla\phi$ has been used. This expression equals zero provided that the quantity $f$ is constant in time, i.e. $df/dt = 0$.