Finite, surjective, and unramified does not imply etale. E.g. suppose that $Y$ is a proper closed subscheme of $X$, and we consider the map $X \coprod Y \to X$ defined as the disjoint union of the identity on $X$, and the given closed immersion $Y \to X$ on $Y$.

Then this map is finite, unramified, and surjective, but not etale. (See Sandor's answer for the missing condition, which is flatness!)

Finite, surjective, and unramified does not imply etale. E.g. suppose that $Y$ is a proper closed subscheme of $X$, and we consider the map $X \coprod Y \to X$ defined as the disjoint union of the identity on $X$, and the given closed immersion $Y \to X$ on $Y$.

Then this map is finite, unramified, and surjective, but not etale. (See Sandor's answer for the missing condition, which is flatness!)

Added: A more interesting example is given by letting $X$ be a nodal cubic, letting $\tilde{X}$ be the normalization, and considering the natural map $\tilde{X} \to X.$ This map is not flat and certainly not etale, but it is unramified. (Formally, each branch through the node maps by a closed immersion into the nodal curve.)