I would like to check if my understanding of the process of finding the rational form of the given matrix is correct.
$$B=\begin{bmatrix}c&0&-1 \\ 0&c&1\\-1&1&c\end{bmatrix}$$
The characteristic function is $$det(xI - B)=\begin{bmatrix}x-c&0&-1 \\ 0&x-c&1\\-1&1&x-c\end{bmatrix}=(x-c)^3$$
The minimal polynomial is also $(x-c)^3$ since $B-cI \not=0$ and $(B-cI)^2 \not=0$. Now the minimal polynomial is $(x-c)^3 = x^3 - 2x^2c + 2xc^2 - c^3$, so the companion matrix is a $3 \times 3$ matrix is the same as the rational form matrix and is equal to:
$$B=\begin{bmatrix}0&0&c^3 \\ 1&0&-2c^2\\0&1&2c\end{bmatrix}$$
Is this correct?
Your calculations and results are correct.
One little remark: You should not name the rational form matrix $B$, since it is not $B$. The matrix $B$ is only similar to its Frobenius normal form.