Jordan normal form of $;begin{pmatrix}0 & 1 & 0 \ 0 & 0 & 1 \ 0 & a & b end{pmatrix},; a,binmathbb{R}$

We can avoid computing generalized eigenvectors. Separate this into $3$ cases.

Case 1: $a = b = 0$. As you said, the matrix is already in Jordan form

Case 2: $a=0$, $b neq 0$. The matrix is upper triangular, so we quickly see that its only eigenvalues are $0$ and $b$. The block associated with $b$ has size $1$, and the because the matrix $M$ has rank $2$, the block associated with $0$ has size $2$.

Now, if $a neq 0$, then $M$ is block upper triangular with the form $$ M = left[begin{array}{c|cc} 0&1&0\ hline 0&0&1\0&a&bend{array}right] $$

Case 3: If $a = -(b/2)^2 neq 0$, then the lower-right block has a repeated eigenvalue, and its Jordan form consists of a single block. Thus, the overall Jordan form has a $0$ followed by a size-$2$ block.

Case 4: In the remaining cases, the lower-right block has eigenvalues that are distinct and non-zero. Thus, $M$ has distinct eigenvalues, which means that its Jordan form is diagonal.

---

As an alternative, it would suffice to note that $M$ is the transpose of the companion matrix associated with the polynomial $p(t) = t^3 - bt^2 - at$. It follows that its Jordan form consists of a single block of maximal size for each of the roots of $p$.