Let $f:X\rightarrow Y$ a faithfully flat morphism between $k$schemes. We assume that the fibers are locally of finite type, do we have that $f$ is locally of finite type?
No. Let $Y$ be $\text{Spec} \mathbb{Z}$. Let $X$ be $\text{Spec} (\mathbb{Z} \times \mathbb{Q})$. $\textbf{Edit}.$ As pointed out, the OP wants an example over a field. As the commenters explain, the same idea works over a field $k$ with $Y = \text{Spec} k[t]$ and with $X=\text{Spec}(k[t] \times k(t))$. 

