What does the notation $\big(\frac{\partial p}{\partial T}\big)_V$ mean? In a book about thermodynamics, I came across the notation shown in the title. A more complete example would be
$$dU = \left(\frac{\partial U}{\partial T}\right)_V\; dT + \left(\frac{\partial U}{\partial V}\right)_T\; dV$$
But I have no idea what the subscript means in this notation, would love some help.
I ran in to trouble with an exercise that said

Assume that gases behave according to a law given by $pV = f(T)$, where $f(T)$ is a function of temperature. Show that this implies $$\left(\frac{\partial p}{\partial T}\right)_V = \frac1V\frac{df}{dT}$$

 A: Ian's comment above is spot-on, but I think it could benefit from some elaboration, so I'll write up a more detailed explanation. We all know that a partial means treat the other variable as constant. It's all about the fact that the other variable could mean any number of things.
When a mathematician thinks of a function, it is a "map" which to each set of values of the argument variables associates a value of the function. So $U = f(T,V)$, for example, defines a value, $U$, for each ordered pair $(T,V)$ in the domain of $f$. But physicists like to write that simply as $U(T,V)$. $U$ here is not a function in the strict mathematical sense: it is a physical quantity whose value happens to depend on $T$ and $V$ (more than that: it is uniquely determined by $T$ and $V$). 
In thermodynamics, the state of a system can be described by different sets of variables. In the simplest situation, some fixed amount (mass) of substance (with uniform properties) can be described by any two of temperature $T$, pressure $p$, volume $V$ and entropy $S$. For any such system, the equations of state relate these to each other and allow one to switch from one pair to another as convenient for a given purpose (just as one can change from rectangular to polar coordinates to specify a point in the plane).
Suppose, for example, that $U(p,V) = c p V$ (this is valid for ideal gas), where $c$ is some constant. What if I wanted instead to find how the same quantity, internal energy, would be expressed in terms of $T$ and $V$? It would be very wrong to simply write $U(T,V) = c T V$ by simply replacing $p$ with $T$ in the formula, as one would with any mathematical function. What is important is to preserve the meaning of $U$ - it's value for a given state of the system - so we use the equation of state to express $p$ in terms of $T$ and $V$ and then substitute. For ideal gas we would get $U = b T$, where $b$ is another constant (because the ideal gas equation of state is $p \propto T/V$). As a function of two variables, U is now actually different!
With the original function, we have
$$ \frac{\partial U}{\partial V} = cp = \left(\frac{\partial U}{\partial V}\right)_p$$
since $p$ is the second variable in that function. But with the $U$ as a function of $T$ and $V$, we'd write
$$ \frac{\partial U}{\partial V} = 0 = \left(\frac{\partial U}{\partial V}\right)_T. $$
This is clearly different from the above.
The same also applies to $p$ itself, if it is expressed in terms of two of the other "basic" variables, as in the specific example in the question. If I have $p = f(T)/V$, then deriving the wanted result is a piece of cake. But if I had considered $p$ as a function of $T$ and $S$, I might get something very different. Suppose $pV = f(T) = KT$ with $K$ a constant (the ideal gas equation of state again). Then
$$
\left(\frac{\partial p}{\partial T}\right)_V = \frac{K}{V} = \frac{p}{T} 
$$
But it can be shown that (I won't get into that here)
$$
\left(\frac{\partial p}{\partial T}\right)_S = \frac{\gamma}{\gamma-1}\frac{p}{T}, $$
where $\gamma$ is the adiabatic index. You can see that the above are not equal for any value of $\gamma$.
In more complex scenarios, a thermodynamic state might not be uniquely defined in only two variables. Then any thermodynamic potential (a function of state) would be a function of maybe three, maybe more variables, and we'd have to write things like
$$  \left(\frac{\partial U}{\partial T}\right)_{V,\; X,\; Y,\; \dots} $$
A: You can verify this here:
$$\left( \frac{\partial p}{\partial T}\right) _V$$
is the partial derivative of $p$ with respect to $T$ while holding $V$ constant.
