Total Derivatives and Thermodynamics Question 1
What are the following quantities functions of: volume, pressure, temperature and mass? For I am very confused when I should be using $d$ or $\partial$ for my derivatives in thermodynamics. 
For example, If I take the total derivative of internal energy [$U(S,V)$] as a function of temperature, I get the following:

$\frac{d}{dT}$U(S,V) = $ \frac{ \partial U}{ \partial T} + (\frac{ \partial U}{ \partial S}) \frac{ds}{dT} + (\frac{ \partial U}{ \partial V}) \frac{dV}{dT}    $

If I multiply both sides by $dT$ to get my desired infinitesimal, I get:

$dU(S,V) = \frac{ \partial U}{ \partial t} dT + (\frac{ \partial U}{ \partial
 S}) ds + (\frac{ \partial U}{ \partial V})dV$

However, my textbook uses the following notation in some questions and I do not think $\partial t$ and $dt$ cancel each other out, but I am not sure

$\frac{d}{dT}$U(S,V) = $ \frac{ \partial U}{ \partial T} + (\frac{ \partial U}{ \partial S}) \frac{ \partial s}{ \partial T} + (\frac{ \partial U}{ \partial V}) \frac{ \partial V}{ \partial T}    $

Question 2
When you turn  to infinitesimal form, which variable are we taking the derivative with respect to? I always assumed time, but my previous question is casting doubles. For example:

$\frac{d}{dt}$U(S,V) = $ \frac{ \partial U}{ \partial t} + (\frac{ \partial U}{ \partial S}) \frac{ds}{dt} + (\frac{ \partial U}{ \partial V}) \frac{dV}{dt}    $

Some authors "cheat" and do the derivative calculations without the numerator. For example:

$F + \Delta F = U + \Delta U - (P + \Delta P)(V + \Delta V)$

How is this connected to a normal calculus (i.e.  total/partial derivatives)?
Summary
I understand kinds types of derivatives (partial/total), but do know know which type thermodynamics uses or when
Previous Research

Partial derivatives vs total derivatives in thermodynamics

 A: If U=U(S,V), then, mathematically, you should be writing $$dU=\left(\frac{\partial U}{\partial S}\right)_VdS+\left(\frac{\partial U}{\partial V}\right)_SdV=TdS-PdV$$Treating S as a function of V and T, you also have $$dS=\left(\frac{\partial S}{\partial T}\right)_VdT+\left(\frac{\partial S}{\partial V}\right)_TdV=\frac{C_v}{T}dT+\left(\frac{\partial S}{\partial V}\right)_TdV$$So, combining these two equations, you have $$dU=C_vdT-\left[P-T\left(\frac{\partial S}{\partial V}\right)_T\right]dV$$
A: Thermodynamics is a minefield for issues like this.
The internal energy, as you say, is a function of two variables - $S$ and $V$.  We can define its partial derivatives with respect to $S$ and $V$ as follows:
$$\left(\frac{\partial U}{\partial S}\right)_V = \lim_{h\rightarrow 0} \frac{U(S+h,V)-U(S,V)}{h}$$
$$\left(\frac{\partial U}{\partial V}\right)_S = \lim_{h\rightarrow 0} \frac{U(S,V+h)-U(S,V)}{h}$$
But now you want to talk about $\frac{\partial U}{\partial T}$, where $T \equiv \left(\frac{\partial U}{\partial S}\right)_V$, and it's not immediately clear what that means, since $U$ is not even a function of $T$.
What we mean is the following: we change $U$ by changing both $S$ and $V$ at the same time:
$$dU = U\big(S+dS,V+dV\big) = \left(\frac{\partial U}{\partial S}\right)_V dS + \left(\frac{\partial U}{\partial V}\right)_S dV $$
Because $T\equiv \left(\frac{\partial U}{\partial S}\right)_V$ is also a function of $S$ and $V$, by changing both $S$ and $V$ at the same time, we also end up changing $T$ :
$$dT = \left(\frac{\partial T}{\partial S}\right)_V dS + \left(\frac{\partial T}{\partial V}\right)_SdV$$
and so we define $\frac{\partial U}{\partial T}$ to be the ratio of these two changes:
$$\frac{\partial U}{\partial T} = \frac{\left(\frac{\partial U}{\partial S}\right)_V dS + \left(\frac{\partial U}{\partial V}\right)_S dV}{ \left(\frac{\partial T}{\partial S}\right)_V dS + \left(\frac{\partial T}{\partial V}\right)_SdV}$$
Now, this quantity is not well-defined because we need to specify precisely how we change $S$ and $V$.  For example, we could ask about $\left(\frac{\partial U}{\partial T}\right)_V$ - the rate of change of the internal energy with respect to $T$ when we hold $V$ constant.  In this case $dV=0$ and we would find
$$\left(\frac{\partial U}{\partial T}\right)_V = \left(\frac{\partial U}{\partial S}\right)_V \big/\left(\frac{\partial T}{\partial S}\right)_V = T \left(\frac{\partial S}{\partial T}\right)_V \equiv c_V$$
where $\left(\frac{\partial S}{\partial T}\right)_V \equiv 1\big/\left(\frac{\partial T}{\partial S}\right)_V$ and we note the right hand side as being simply the definition of the specific heat at constant volume $c_V$.
If you choose some other way to change $S$ and $V$, the result will be different.  For example, we could choose not to keep $V$ constant, but rather the pressure $P \equiv -\left(\frac{\partial U}{\partial V}\right)_S$.
A: Your first equation represents the differential of U with respect to T, as shown here: https://en.wikipedia.org/wiki/Differential_of_a_function.
This function represents a linearization about a specific point in the multi-variable equation, and I seriously doubt that it is correct to multiply that equation by dT in order to eliminate it.
