Deriving partial differential equation for temperature using conservation laws and constituent relations I want to derive the partial differential equation for the temperature T as
$$
\partial_t T(t,x) + v(t,x)\cdot \nabla T(t,x)=0.
$$
Quick note: I'm doing a mathematics major, and so I do not really understand the physical context in depth, I'm trying though.
I have the following information and constituent relations:
$\sigma = -p\;\text{Id }\quad$We are working non-viscous,
$p=\hat{p}(\rho)$, the pressure only depends on the density,
$v(t,x)$, the velocity field of the air,
$f=\hat{f}(\rho, T)$, the density of free energy,
$u=\hat{u}(\rho, T)$, the (density of ?) internal energy,
$s=\hat{s}(\rho, T)$, the entropy,
$q=0$, heat flux is non-existent,
$h=0$, there are no exterior heat sources.
I am given the following constituent relations:
$\hat{p}(\rho)=\rho^2\partial_{\rho}\hat{f}(\rho, T)$,
$\hat{s}(\rho, T)=-\partial_T \hat{f}(\rho, T)$,
$\hat{u}(\rho, T)=\hat{f}(\rho, T) + T\hat{s}(\rho, T)$.
I should also assume constant heat capacity, $\partial_T \hat{u} = c$.

I worked out that $\hat{f}(\rho, T)=f_1(\rho)+f_2(T)$, meaning it has an additive structure.
Also $\partial_{\rho}\hat{u}=\partial_{\rho}\hat{f}$.
Constant heat capacity means $$\partial_{\rho}\partial_{T}\hat{u}=\partial_\rho \big(\partial_T \hat{f}(\rho, T) - \hat{f}(\rho, T) - T\partial_T\hat{f}(\rho, T)\big)
= \partial_\rho \big(\partial_T f_2(T) - \hat{f}(\rho, T) - T\partial_T f_2(T)\big)
= \partial_{\rho} \hat{f}(\rho, T) = \partial_{\rho} f_1(\rho)
= 0,
$$
and thus also 
$$\partial_{\rho} \hat{u} = \partial_{\rho} \hat{f} = \partial{\rho} f_1(\rho) = 0.$$
I think I'm making a mistake here, because if $\partial_{\rho} \hat{f} = 0$, then also $\hat{p}=\rho^2 \partial_{\rho} \hat{f} = 0$, which seems like a weird conclusion. Then also $\sigma=\hat{p}\;\text{Id}=0$, does that seem right?
It then says I should use the conservation laws for mass, momentum and energy, which are not in the exercise, but which I got from the book "Mathematical Modeling" by Christof Eck, Harald Garcke, Peter Knabner:
Mass: with $\varrho$ a "mass density" and $v$ a velocity field
$$
\frac{d}{dt} \int_{\Omega(t)} \varrho(t, x)dx = 0,
$$
which, using Reynolds transport theorem, leads to the possibly more useful
$$
\partial_t \varrho + \nabla\cdot(\varrho v) = 0.
$$
Linear momentum:
$$
\frac{d}{dt} \int_{\Omega(t)} \varrho(t,x)v(t,x)dx = \int_{\Omega(t)} \varrho(t,x)f(t,x)dx + \int_{\partial\Omega(t)} b(t,x)ds_x,
$$
where $f$ is a force density (is that the same as $f$ above for free energy?), $b$ is a force density onto the surface.
This can also be written as
$$
\varrho(\partial_t v + (v\cdot \nabla)v)-\nabla\cdot\sigma=\varrho f.
$$
Energy:
$$
\frac{d}{dt} \int_{\Omega(t)} \big(\varrho(\frac{|v|^2}{2} + u)\big)dx = \int_{\Omega(t)} \varrho f \cdot v dx + \int_{\partial \Omega(t)} \sigma n\cdot v ds_x.
$$
which can also be written
$$
\varrho\partial_t u + \varrho v \cdot \nabla u - \sigma : D_v = 0,
$$
where the colon is the matrix inner product.
Now trying to plug in our constituent relation $\hat{u}=\hat{f}+T\hat{s}$ in this rewritten energy equation, we would get a term like $\partial_t \hat{f}$, but $f$ is a function of $\rho$ and $T$ and I do not see how that matches.
Now I really don't know how to use all this information. The conservation laws all use functions with arguments time and space, while the constituent relation functions are all of density and temperature. I do not understand how to link that.

EDIT:
By writing and rereading the post I realised something:
I realised that what they probably mean is that $\rho$ and $T$ are both functions of $x$ and $t$, so then we get that $\hat{u}(\rho(x,t),T(x,t))$ are really also functions of $x$ and $t$ and in this way we can plug the constituent relations into the conservation laws using the chain rule (though I don't know how to manage all the different terms, it gets a little complex)
 A: First of all, what was said in the question and assumed wrong:

Constant heat capacity means $$\partial_{\rho}\partial_{T}\hat{u}=\partial_\rho \big(\partial_T \hat{f}(\rho, T) - \hat{f}(\rho, T) - T\partial_T\hat{f}(\rho, T)\big)
= \partial_\rho \big(\partial_T f_2(T) - \hat{f}(\rho, T) - T\partial_T f_2(T)\big)
= \partial_{\rho} \hat{f}(\rho, T) = \partial_{\rho} f_1(\rho)
= 0,
$$

is indeed wrong, because it should read
$$\partial_{\rho}\partial_T \hat{u} = \partial_{\rho}(\partial_T \hat{f} - \mathbf{\partial_T} \hat{f} - T\partial_T\partial_T \hat{f}),$$
from which nothing useful can be obtained for this problem.

To solve this we only need the conservation of mass and energy, and a lot of chain rules.
First of all, we can use the $\rho$ inside the heat equation as if it was $\varrho$ since it is a mass density.
We will use the conservation of energy as a guideline and first simplify this equation
$$
0 = \rho\partial_t u + \rho v\cdot \nabla u - \sigma : Dv\\
= \rho\big( \partial_t u + v\cdot \nabla u - \rho \partial_{\rho} f \cdot \text{Id} : Dv\big),
$$
so we want to solve
$$
\partial_t u + v\cdot \nabla u - \rho \partial_{\rho} f \cdot \text{Id} : Dv = 0,
$$
which we will do by rewriting all three summands and then plugging them back in.
First summand: Using $\partial_\rho u = \partial_\rho f$, constant heat capacity ($\partial_T u = c$) and the chain rule, we get
$$
\partial_t u = \partial_\rho u \partial_t \rho + \partial_T u \partial t T = \partial_\rho u \partial_t \rho + c\partial_t T = \partial_\rho f \partial_t \rho + c\partial_t T.
$$
Second summand: Using again the chain rule, constant heat capacity ($\partial_T u = c$) and the definition of the spacial gradient, we get
$$
v\cdot \nabla u = \sum_{i=1}^3 v_i \cdot \partial_{x_i} u \\
= \sum_{i=1}^3 v_i \cdot \big(\partial_\rho u \partial_{x_i} \rho + \partial_T u \partial_{x_i} T\big)\\
= \sum_{i=1}^3 \big(v_i \cdot \partial_\rho u \partial_{x_i} \rho\big) + \sum_{i=1}^3 \big(v_i \cdot \partial_T u \partial_{x_i} T\big)\\
= \partial_\rho u \sum_{i=1}^3 \big(v_i \cdot \partial_{x_i} \rho\big) + c\sum_{i=1}^3 \big(v_i \cdot \partial_{x_i} T\big)\\
= \partial_\rho u (v \cdot \nabla \rho) + cv\cdot\nabla T.
$$
Third summand: Using that ":" is the matrix inner product, $Dv$ is the total derivative of $v$, and $\nabla \cdot$ is the divergence we can rewrite the third summand as
$$
\rho\partial_\rho f \cdot \text{Id} : Dv = \rho \partial_\rho f \sum_{i=1}^3 \partial_{x_i} v_i = \rho \partial_\rho f \nabla\cdot v.
$$
As posted in the question, from the mass equation one can derive the continuity equation $\partial_t \rho + \nabla\cdot (\rho v) = 0$. Using the definitions of the divergence and gradient we find
$$
\nabla \cdot (\rho v) - (\nabla \rho) \cdot v = \rho(\nabla \cdot v).
$$
Plugging this into our already rewritten third summand, and again using $\partial_\rho f = \partial_\rho u$, we get
$$
\rho \partial_\rho f \cdot \text{Id} : Dv = \rho \partial_\rho f \nabla \cdot v = \partial_\rho f (\nabla \cdot (\rho v) - (\nabla \rho) \cdot v) = \partial_\rho f (-\partial_t \rho - (\nabla \rho) \cdot v) 
\\= - \partial_\rho f \partial_t \rho - \partial_\rho f (\nabla \rho \cdot v)
= - \partial_\rho f \partial_t \rho - \partial_\rho u (\nabla \rho \cdot v).
$$
Now plugging everything back into the energy conservation equation we get
$$
0 = 
\partial_t u + v\cdot \nabla u - \rho \partial_{\rho} f \cdot \text{Id} : Dv
\\=
\partial_\rho f \partial_t \rho + c\partial_t T
+
\partial_\rho u (v \cdot \nabla \rho) + cv\cdot\nabla T
- \partial_\rho f \partial_t \rho - \partial_\rho u (\nabla \rho \cdot v)
\\=
c(\partial_t T + v\cdot\nabla T)
$$
and thus we get
$$
\partial_t T + v\cdot\nabla T = 0.
$$
