$f=f(x(p,q),y(p,q),z(p,q))$, where $p$ and $q$ are independent.
Does this mean that
$\frac{\partial f}{\partial x}=\frac{\partial f}{\partial p}\frac{\partial p}{\partial x}+\frac{\partial f}{\partial q}\frac{\partial q}{\partial x}$, where we suppose that $p$ and $q$ are invertible functions of $x$?...
How do we express $\frac{\partial f}{\partial x}$?
On
The chain rule would yield $$\dfrac{\partial f}{\partial p} = \dfrac{\partial f}{\partial x} \, \dfrac{\partial x}{\partial p} + \dfrac{\partial f}{\partial y} \, \dfrac{\partial y}{\partial p} + \dfrac{\partial f}{\partial z} \, \dfrac{\partial z}{\partial p} \, , \qquad \dfrac{\partial f}{\partial q} = \dfrac{\partial f}{\partial x} \, \dfrac{\partial x}{\partial q} + \dfrac{\partial f}{\partial y} \, \dfrac{\partial y}{\partial q} + \dfrac{\partial f}{\partial z} \, \dfrac{\partial z}{\partial q} \, .$$
I think this is a problem that comes up with sloppy notation on the part of multivariable calculus techniques in general. They want $f:\mathbb R^2\rightarrow\mathbb R$ expressed as a composition of two functions: one from $\mathbb R^2$ to $\mathbb R^3$, whose inputs are $(p, q)$ and outputs are $(x(p,q), y(p,q), z(p,q))$; the second, from $\mathbb R^3$ to $\mathbb R$. Where I think it is sloppy is that they are also calling this intermediate function $f$.
I would write $f(p,q) = g(x(p,q), y(p,q), z(p,q))$. That way we have $f: \mathbb R^2\rightarrow \mathbb R$ whose inputs are $(p, q)$, distinct from $g: \mathbb R^3 \rightarrow\mathbb R$ whose inputs are $(x, y, z)$. But the notation is shorter the way they put it, and Nunya's answer works if you understand what's going on.
So for this problem $\partial f/\partial x$ is its own expression, nothing to do with $p$ and $q$; in my notation that would be $\partial g/\partial x$. And the Chain Rule would say for example,
$$ \frac{\partial f}{ \partial p} = \frac{\partial g}{\partial x} *\frac{\partial x}{\partial p} + \frac{\partial g}{\partial y} * \frac{\partial y}{\partial p} + \frac{\partial g}{\partial z} * \frac{\partial z}{ \partial p}. $$
Alternatively, rename the composition some other function, and keep the intermediate as $f$. But referring to $f$ so ambiguously doesn't help even though it's basically standard practice.
As for invertible functions in multivariable calculus, a warning: don't casually try to flip derivatives like that. For one thing, a function $x(p, q)$ is not invertible because its input (meaning domain) is some part of $\mathbb R^2$ and its output is something in $\mathbb R$. Second, even if you have an invertible function $\mathbb R^n\rightarrow\mathbb R^n$, to invert the derivative you need linear algebra (inverse of a $n\times n$ matrix) not just taking reciprocals, and you can't do it for each variable separately.