Yes.

Take $f=z^2+(x^3+y^3)(y^3+x^4)$. To compute the resolution, consider the projection $U\to {\mathbb C}^2$ onto the $x,y$ coordinates: it is a double cover branched on the curve $B:=\{(x^3+y^3)(y^3+x^4)=0\}$. Blow up the origin in ${\mathbb C}^2$, and let $U'$ be the surface obtained by base change and normalization: $U'$ is a double cover branched on the strict transform $B'$ of $B$ and it is smooth since $B'$ is smooth. The exceptional locus of the resolution $U'\to U$ is the inverse image $Z\subset U'$ of the exceptional curve of the blow up. It is a standard computation to show that $Z$ is irreducible with $Z^2=-2$, $p_a(Z)=2$ and $Z$ has a simple cusp as singularity. So the resolution is minimal and $Z$ is not normal crossings.