Existence of MAC implies existence of OWF
It is not hard to construct a secure MAC where $f(k) = \textsf{MAC}(k,0)$ is not one-way. Take any secure $\textsf{MAC}$ and define a new one as 

$$\textsf{MAC}^*\Bigl( (k_1,k_2), m \Bigr) = \textsf{MAC}(k_1,m) \oplus k_2.$$

It should be relatively clear that $\textsf{MAC}^*$ is also a secure MAC. Now given arbitrary $m$ and $t$, it is easy to find a key under which $\textsf{MAC}^*((k_1,k_2,m)=t$. Just choose arbitrary $k_1$ and then solve for $k_2 = t \oplus \textsf{MAC}(k_1,m)$. This leads to a natural way to invert the $f$ function you propose.

Another way to look at things: The function $f(k) = \textsf{MAC}(k,0)$ can only rely on the fact that the MAC is secure against an attacker who sees just one output. But there are information-theoretic MACs with this property, and you can't expect them to imply a OWF.

To get around this, you should use $f(k) =\textsf{MAC}(k,0) \| \textsf{MAC}(k,1) \| \cdots \| \textsf{MAC}(k,\ell)$ as your OWF candidate, where $\ell$ depends on the key size of this MAC. But even then the argument of one-wayness is not trivial.