This is Corollary 6.2.3 in Liu's book.
Let $f: X \to S$ be a morphism of finite type of locally Noetherian schemes. Then $f$ is unramified at a point $x \in X$ if and only if $\Omega_{X/S, x} = 0$.
Relevant definitions:
$f: X \to Y$ is unramified at $x \in X$ if $k(y) \to k(x)$ is separable and $m_y \mathcal{O}_{X,x} = m_x$ where $y = f(x)$.
For an algebraic variety $X$ over a field $k$, say that it is smooth at $x \in X$ if the points of $X_\overline{k}$ lying above $x$ are regular points of $X_\overline{k}$.
Here is Liu's proposed proof of the corollary:
Denote $\Omega_{X/S}$ by $\Omega_X$. Let $s = f(x)$. Then $\Omega_{X,x} \otimes k(s) = \Omega_{X_s,x}$. The morphism is unramified at $x$ iff $X_s$ is smooth at $x$ and $x$ is an isolated point in $X_s$.
The last sentence in the above proof seems to be the crux of the rest of the proof. For the justification, he cites the following result:
Let $f: X \to Y$ be a morphism of finite type between locally Noetherian schemes. Then $f$ is unramified if and only if for every $y \in Y$, the fiber $X_y$ is finite, reduced, and if $k(x)/k(y)$ is separable for every $x \in X_y$.
The above tells me that unramified is equivalent to the fiber being finite and reduced. However, Liu claims it is equivalent to the fiber being smooth at the point, which is moreover an isolated point. There is clearly a step missing here.
The next part of the proof says that
If $f$ is unramified at $x$, then $\Omega_{X_s,x} = 0$.
So $$\Omega_{X_s,x} = \Omega_{\mathcal{O}_{X_s,x}/\mathcal{O}_{X,x}} = \Omega_{\mathcal{O}_{X,x} \otimes k(s) / \mathcal{O}_{X,x}}$$
I don't see how unramified makes that zero.
This proof is a complete mess and I've been searching online for alternative proofs. The Stacks Project seems to just define unramified as this but I could not find where they show the equivalence. If anyone can provide an alternative proof via a (preferably free) resource, I would appreciate that.