Given a secure PRF $f(k, x)$, is $f(x, k)$ also a secure PRF? Let $f: \{0,1\}^n \times \{0,1\}^n \rightarrow \{0,1\}^n$ be a secure PRF.
Define $F(k, x) = f(x, k)$.
Is $F(k, x)$ also a secure PRF?
No. A PRF assumes that the key is chosen randomly.

Example based on @LuisCasillas's comment: With a secret random key, you cannot determine the key from input output pairs. If you reverse the role of key and input, then you have the question: Can you learn "input" ($k$) from output ($f(x,k)$) and "key" ($x$)?

Yes. If $F(k, x) = f(x, k) = E_x(k)$, where $E$ is a PRP, then you can recover $k$ from one input/output pair of $F$. If your output is $c = E_x(k)$ then anyone can compute $k$ if they know $c$ and $x$. 
$$k = E^{-1}_x(c) = E^{-1}_x(E_x(k))$$

(Where $E$ is block encryption and $E^{-1}$ is the inverse, decryption.

This answers your question because a PRP is a PRF because a permutation is a function. (Or maybe I'm wrong on that. There should be a caveat for small blocksizes. The presence or absence of collisions could distinguish a PRP from a non-PRP PRF.)

However PRPs are an example, not the exception. It's still not safe in general. There can be chosen key attacks, related key attacks, weak key sets, etc. A specific algorithm with no weakness when used correctly is not guaranteed to also be free of weakness when used incorrectly.