Iterating Total Derivative How to. Let $f =f(x(t),y(t))$
Then the total derivative of f with respect to t is given by:
$$\frac {}{} = \frac {}{} + \frac {}{}$$
Say still we have  $f=f(x(t),y(t))$. What is then $\frac {\partial^2 f}{ \partial t^2} $? Is it just $$ \frac {\partial f}{ \partial x} \frac {d^2x}{dt^2}+ \frac {\partial f }{\partial y} \frac {d^2y}{dt^2} $$ ? (Just by applying chain rule again?)
Also, this seems to be related to the Jacobian  J=J(f). Is this correct? If so, can we define $\frac{\partial^2 f}{ \partial t^2}$ as $JJ^T$ , where $J^T$ is the transpose of $J$?
 A: First, if $f$ is only a function of $x$ and $y$, then partial derivative over $t$ is $0$, so I'll asume we are talking about total derivative.
Just do it in steps. First derivative:
$$\frac{df}{dt}=\frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt}$$
Now differentiate again. Each term will have to be differentiated as a product.
$$\frac{d^2 f}{dt^2}=\frac{d}{dt}\left(\frac{\partial f}{\partial x}\right)\frac{dx}{dt}+\frac{\partial f}{\partial x}\frac{d^2x}{dt^2}
+\frac{d}{dt}\left(\frac{\partial f}{\partial y}\right)\frac{dy}{dt}+\frac{\partial f}{\partial y}\frac{d^2y}{dt^2}=$$
$$=\left(\frac{\partial^2 f}{\partial x^2}\frac{dx}{dt}+\frac{\partial^2 f}{\partial x\partial y}\frac{dy}{dt}\right)\frac{dx}{dt}+\frac{\partial f}{\partial x}\frac{d^2x}{dt^2}
+\left(\frac{\partial^2 f}{\partial y\partial x}\frac{dx}{dt}+\frac{\partial^2 f}{\partial y^2}\frac{dy}{dt}\right)\frac{dy}{dt}+\frac{\partial f}{\partial y}\frac{d^2y}{dt^2}$$
If you want to write it in a nicer, more symmetric form, introduce the following objects. The Hessian matrix:
$$H=(\partial_i \partial_j f)=\begin{bmatrix}
\frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x\partial y}\\
\frac{\partial^2 f}{\partial x\partial y} & \frac{\partial^2 f}{\partial y^2}\\
\end{bmatrix}$$
The derivatives of the trajectory vector $\vec{r}=(x,y)$:
$$\dot{\vec{r}}=\left\lbrace\frac{dx}{dt},\frac{dy}{dt}\right\rbrace,\quad \ddot{\vec{r}}=\left\lbrace\frac{d^2x}{dt^2},\frac{d^2 y}{dt^2}\right\rbrace$$
The gradient of the function:
$$\nabla f=\left\lbrace\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}\right\rbrace$$
Then you can write down the derivative as such:
$$\frac{d^2 f}{dt^2}=\dot{\vec{r}}H\dot{\vec{r}}+(\nabla f)\cdot \ddot{\vec{r}}$$
You may interpret $\dot{\vec{r}}$ as velocity, $\ddot{\vec{r}}$ as acceleration (these are properties of your trajectory through the domain of the field $f$). The other two objects are properties of the field (they are defined at every point, no matter if your trajectory passes through). The Hessian talks about the "curvature" of the field (second derivatives) and the gradient is the slope of steepest descent.
The second derivative of the field "felt" by the traveling particle (imagine feeling height above sea level) is composed of two terms. The first term describe how much the function value $f(x,y)$ "curves" (how much second derivative it has) in the direction of travel (told by velocity). The second term describes how much second derivative you feel just because you are accelerating downhill. Observe that you missed the first term in your question - you did not take into account that the landscape itself can look like a parabola in cross section and thus have a second derivative even if your velocity is constant (imagine $x=kt$, $y=0$, $f=x^2+y^2$. Your formula gives 0, even though $f=k^2t^2$ clearly has second derivative $2k^2$).
The Jacobian is a different kind of object that comes into play when you are changing variables (mapping one space to another), which is not the case here.
