If $L/K$ is a finite separable fields extension, then it is formally smooth, i.e., for every commutative diagram
where $B$ is a ring and $I\subseteq B$ is an ideal satisying $I^{2}=0$ there exists a lifting $\lambda\colon L\to B$ making the diagram commute.
I need to prove the converse. If $L/K$ is finite and not separable, then it is not formally smooth. This is how one could proceed. If $K$ has characteristic $p$ and $a\in L$ is not a $p$-th power in $K$ then suppose that $K(\sqrt[p]{a})$ is formally smooth over $K$.
Then $$K(\sqrt[p]{a})\otimes_{K} K(\sqrt[p]{a})\cong \frac{K(\sqrt[p]{a})[x]}{(x^{p}-a)}\cong\frac{K(\sqrt[p]{a})[x]}{(x-\sqrt[p]{a})^{p}}$$ should be formally smooth because https://stacks.math.columbia.edu/tag/00TJ but I want to prove that this is a contradiction because, in reality, $\frac{K(\sqrt[p]{a})[x]}{(x-\sqrt[p]{a})^{p}}$ is not formally smooth. But I am unable to prove this.
To conclude the proof, one could demonstrate that if $L/K$ were formally smooth, then by considering $K_{0}$ such that $K\subseteq K_{0}\subseteq L$ and $L=K_{0}(\sqrt[p^{r_{1}}]{a_{1}},\dots,\sqrt[p^{r_{m}}]{a_{m}})$ we get a formally smooth subextension $K_{0}(\sqrt[p](a_{1}))/K_{0}$, so a contradiction. This is another point that I am unable to prove. Is it true that given $K\subseteq F \subseteq L$, if $L/K$ is formally smooth, then both $L/F$ and $F/K$ are also formally smooth?
You want the following statement: Let $K$ be a characteristic $p$ field. Then $K[x]/x^p$ is not formally smooth over $K$.
Let $B=K[\epsilon]/\epsilon^{p+1}$ and $B/I=K[\epsilon]/\epsilon^{p}$. Let $K\to B$ be the obvious map, and let $K[x]/x^p\to B/I$ send $x$ to $\epsilon$. Then there is no lift $K[x]/x^p\to B=K[\epsilon]/\epsilon^{p+1}$, since $\alpha^p=0$ in $B$ implies $\alpha\in\epsilon^2B$.