Let $f:X \rightarrow Y$ be a finitely presented separated etale morphism, with $Y$ quasicompact and quasiseparated.
By Zariski’s main theorem, we can factor $f$ as $f= g \circ j$ with $j$ an open immersion and $g$ finite.
Can we choose $g$ to be etale?
You can not expect there to be an etale extension. Here is one example.
Consider the morphism $f:\overline{X}:=\mathbb{A}^1\to \mathbb{A}^1=:Y$ defined by
$$f(x) = x^2(x-1).$$
Note that $f$ only ramifies at $0$ and $2/3$ in $\overline{X}$. Define $X:=\overline{X}\setminus\{0, 2/3\} = \mathbb{A}^1\setminus \{0, 2/3\}$. Note that $$f|_X: X\to Y$$ is surjective, flat, finite type, quasi-finite and even etale. (We just removed the ramification point.)
We can not extend $f|_X:X\to Y$ to a finite etale morphism $X'\to Y$.
If $X'\to Y$ is a finite etale morphism and $X\to X'$ is an open immersion, then $X'\cong\overline{X} =\mathbb{A}^1$. The extension $X'\to Y$ has to equal the morphism $f:X\to Y$ (by separatedness of $X$ and $Y$), and is therefore not etale.
