A partial answer: a result of Juraj Bosák says that if all vertices of a graceful graph have even degree, then the graph has $4k$ or $4k+3$ edges for some integer $k$. (Proof added below.)


Since $K_5^{(3)}$ has $30$ edges, it is ungraceful.

Gallian's survey attributes this result to Rosa's 1967 paper _On certain valuations of the vertices of a graph_.
See also Don Knuth's work-in-progress section 7.2.2.3 of _The Art of Computer Programming_ (https://www-cs-faculty.stanford.edu/~knuth/fasc7a.ps.gz) for a discussion of computational aspects of graceful labelings as constraint satisfaction problems.

**Lemma 7.2.2.3O from _The Art of Computer Programming_.** _In any graceful labeling of a graph with $4k+1$ or $4k+2$ edges, the number of vertices with an odd degree and an odd label is always odd._

**Proof.** We have $\sum_{uv\in E(G)}|l(u)-l(v)|=1+2+\dots+m=\binom{m+1}{2}$ when there are $m$ edges; and a given vertex $v$ appears exactly $\deg(v)$ times in this sum. Working modulo $2$, we also have $|l(u)-l(v)|\equiv l(u)+l(v)$. Therefore $\sum_v\deg(v)l(v)\equiv\binom{m+1}{2}$. But $\binom{m+1}{2}\equiv1$ when $m=4k+1$ or $m=4k+2$. $\square$