Is this a Fixed Parameter Tractable algorithm?
First and foremost, I would advise having another look at the basic definitions. An FPT instance is identified by a string $x$ and an integer $k$ such that $(x, k)$ is a yes-instance. 

Put briefly, if $f(k)$ is not polynomial in terms of $n$ and still depends on $n$, then you cannot obtain an FPT algorithm. We require that $k$ is fixed, or independent of $n$, and thus $f(k)$ is always independent of $n$. However, if $f(k) = n^{O(1)}$, then our FPT algorithm is in fact a polynomial time algorithm. Hence, if $f(k) = O(2^n)$, the answer is definitely no.