One can construct such an example artificially. Denote the key to the hash function by $s$ (and recall that this key is public). Now, let $H$ be any universal one-way hash function, and define $H'_s(s) = H'_s(s+1)$. This requires changing just one point in $H$, and so is easy.
Notice that $H'$ is clearly not collision resistant: given $s$, just output the pair $(s,s+1)$. However, I argue that $H'$ is still a universal one-way hash function (often called target collision-resistant). The reason for this is that the only way that this change to $H$ can help is if the adversary chooses $x=s$ or $x=s+1$ before the key is chosen. However, this can only happen with negligible probability, unless the key space is polynomial in size. However, if the key space is so small, then the hash function cannot be secure anyway (basically, in such a case, you can just get rid of the key).