Partial derivative in hydrostatic equilibrium in star In a simple model, a gaseous, non-rotating star consists of many thin, concentric spherical shells with radius $r$ and mass $\text{d}m$. The total mass of the shells within radius $r$ is $m$. The model also describes the evolution of the star in time $t$. At each shell we can evaluate various physical quantities, e.g. the density $\rho$ as a function of the independent variables $(r,t)$.
It is often convenient to replace the independent 'Eulerian' variables $(r,t)$ with the independent 'Lagrangian' variables $(m,t)$. What is the expression for the partial derivative of a quantity (e.g. $\rho$) with respect to $t$?
Kippenhahn, Weigert and Weiss (2012, page 5) give $$\left(\frac{\partial}{\partial t}\right)_m=\frac{\partial}{\partial r}\cdot\left(\frac{\partial r}{\partial t}\right)_m+\left(\frac{\partial}{\partial t}\right)_r$$ where the subscipted variables are constant.
Question 1: how can the validity of this expression be demonstrated mathematically?
Question 2: can the partial derivative of $r$ with respect to $t$ at constant $m$ be interpreted physically as the 'radial speed' of the gas in a shell?
 A: 
Question 1: how can the validity of this expression be demonstrated mathematically?

Suppose we have two functions, $F = F\left( t, r\left( t \right) \right)$ and $G = G\left( t, m\left( r \right) \right)$, then the total derivative operator with respect to time can be expressed as:
$$
\begin{align}
  \frac{ d }{ d t } & = \left( \frac{ \partial }{ \partial t } \right)_{r} + \frac{ d r }{ d t } \left( \frac{ \partial }{ \partial r } \right)_{t} \tag{0a} \\
  \frac{ d }{ d t } & = \left( \frac{ \partial }{ \partial t } \right)_{m} + \frac{ d m }{ d t } \left( \frac{ \partial }{ \partial m } \right)_{r} \tag{0b}
\end{align}
$$
Since we know that $r = r\left( t \right)$ (i.e., $r$ is only explicitly dependent on $t$), then we can rewrite $\tfrac{ d r }{ d t }$ as the partial derivative.  Further, since $r$ does not explicitly depend upon $m$, it is assumed that $m$ is constant in the partial derivative (this is always true of partial derivatives, i.e., the variable with which differentiation is being perform is the only one considered to vary during the application of that operator).
We also assume that mass is neither created or destroyed at any point here, so the total derivative of $m$ should be zero.  If we then equate Equations 0a and 0b after the proper substitutions, then we can show that:
$$
\left( \frac{ \partial }{ \partial t } \right)_{m} = \left( \frac{ \partial }{ \partial t } \right)_{r} + \left( \frac{ \partial r }{ \partial t } \right)_{m} + \left( \frac{ \partial }{ \partial r } \right)_{t} \tag{1}
$$

Question 2: can the partial derivative of $r$ with respect to $t$ at constant $m$ be interpreted physically as the 'radial speed' of the gas in a shell?

The short answer is yes.  The reason why is explained below.
Going back to the description of the star, each shell has a mass of $dm$ and let's assume each shell has a thickness of $dr$.  The radius doesn't explicitly depend on any variable except time.  If each shell has the same thickness and total mass and we assume a static shell, i.e., $\left( \tfrac{ \partial r }{ \partial t } \right)_{m} = 0$, then Equation 1 simply states that the quantity in question at some radius $r$ and mass $m$ must be constant in time (which should be obvious from the static assumption).
So to look at a more interesting example, let's consider the quantity of mass density, $\rho$, and not assume the gas is static.  Then Equation 1 can be interpreted as follows:

*

*$\left( \tfrac{ \partial \rho }{ \partial t } \right)_{m}$ is the variation of the mass density with respect to time at some constant mass.  If we take a scenario where the gas in the star fluctuates, then this expression represents the time variation of the mass density in a varying volume (i.e., radial location can vary) to ensure a constant total mass (assuming spherical symmetry, of course).

*$\left( \tfrac{ \partial \rho }{ \partial t } \right)_{r}$ is the variation of the mass density with respect to time at some constant radius.  Again for the fluctuating gas example, this is more easy to imagine.  It's just the time variation of the mass density at some radial location.  If the fluctuations do not alter the mass density, this will be zero.  If they do alter the mass density (e.g., compressive gas), then this term will not be zero.  Note that this term is clearly not equivalent to $\left( \tfrac{ \partial \rho }{ \partial t } \right)_{m}$ under such conditions.

*In the fluctuating gas expression, we now see that the last terms are physically explained as follows.  The $\left( \tfrac{ \partial r }{ \partial t } \right)_{m}$ is the fluctuation velocity of the gas and $\left( \tfrac{ \partial \rho }{ \partial r } \right)_{t}$ is the radial density gradient at some arbitrary time $t$.  If there is a radial gradient in $\rho$ (which there will be for a real star) and any sort of flow of the gas, then these terms can be non-zero and then the partial time derivatives of $\rho$ with constant mass and radius will not be equivalent.

A: After further reading in Kippenhahn, Weigert and Weiss, I try to answer my question 1. This answer seems equivalent to honeste_vivere's answer, but different in the assumptions.
From the chain rule for two independent variables, equation (14.5.3):
suppose x=g(u,v) and y=h(u,v) are differentiable
functions of u and v, and z=f(x,y) is a differentiable function
of x and y. Then, z=f(g(u,v),h(u,v)) is a differentiable function
of u and v, and
$$\frac{\partial z}{\partial u}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial u}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial u}$$
Applying this for example to $\rho=\rho(r,t)$ with $r=r(m,t)$ depending on both $m$ and $t$ (as opposed to $r=r(t)$ in honeste_vivere's assumption), we have at constant $m$:
$$\left(\frac{\partial\rho}{\partial t}\right)_m = \left(\frac{\partial\rho}{\partial r}\right)_t \left(\frac{\partial r}{\partial t}\right)_m + \left(\frac{\partial\rho} {\partial t}\right)_r\left(\frac{\partial t}{\partial t}\right)_m$$
Since the rightmost partial derivative equals 1, this reduces to
$$\left(\frac{\partial\rho}{\partial t}\right)_m = \left(\frac{\partial\rho}{\partial r}\right)_t \left(\frac{\partial r}{\partial t}\right)_m + \left(\frac{\partial\rho} {\partial t}\right)_r$$
Thanks to electronpusher for referring to the multivariable chain rule, which I didn't know.
