Your approach seems correct. Namely, near a breakaway point two real poles should "meetup" and become a pair of complex conjugate poles or vice versa, thus at a breakaway point one would have repeated (real) poles. Repeated poles also means that the derivative of the characteristic equation with respect to $s$ at the value of the pole should be equal to zero. For example define the general openloop as
$$
H_{ol}(s) = k \frac{N(s)}{D(s)}, \tag{1}
$$
with $N(s)$ and $D(s)$ polynomials in $s$. Thus the characteristic equation would be
$$
D(s) + k\,N(s) = 0. \tag{2}
$$
The derivative of the characteristic equation would be
$$
D'(s) + k\,N'(s) = 0, \tag{3}
$$
with $X'(s)$ denoting the derivative of $X(s)$ with respect to $s$. However, inorder to assure that $s$ is a pole of $(2)$ one can indeed use
$$
k = -\frac{D(s)}{N(s)}. \tag{4}
$$
It can be noted that the $k$ in $(2)$ should not change value and thus do not contribute additional terms to the derivative in $(3)$. Substituting $(4)$ into $(3)$ yields
$$
D'(s) - \frac{D(s)}{N(s)}N'(s) = \frac{D'(s)\,N(s) - D(s)\,N'(s)}{N(s)} = 0. \tag{5}
$$
Since $N(s)$ is a polynomial $(5)$ is equivalent to
$$
D'(s)\,N(s) - D(s)\,N'(s) = 0. \tag{6}
$$
Even though it is stated that $k$ should remain constant, it can be shown that $(6)$ is equivalent to setting the derivative of $k$ with respect to $s$ to zero. Namely, by using the quotient rule it can be shown that the derivative of $k$ is equal to
$$
k' = -\frac{D'(s)\,N(s) - D(s)\,N'(s)}{N(s)^2}. \tag{7}
$$
When using again that $N(s)$ is a polynomial yields that setting $(7)$ to zero is equivalent to $(6)$.
Substituting your openloop transfer function does indeed yield
$$
2s^3+18s^2+48s+38=0. \tag{8}
$$
The only thing I could remark on is that you incorrectly rounded the corresponding roots of $(8)$. For example the first root rounded to two decimal places should be $s=-1.47$.
It can be noted that the other roots of $(8)$ are also breakaway points. Only, if you substitute those values into $(4)$ yields negative values for $k$, while normally a rootlocus plot considers only $k\in [0,\infty)$.
The closed-loop transfer function is,
\begin{align}
y &= G(x-z) \\
&= G(x-Hy) \\
(1+GH)y &= Gx \\
y &= \frac{G}{1+GH}x
\end{align}
Let $n_G$, $n_H$ be the numerator of $G$ and $H$, and $d_G$, $d_H$ be the denominator of $G$ and $H$.
\begin{align}
y &= \frac{G}{1+GH}x \\
&= \frac{\frac{n_G}{d_G}}{1+\frac{n_G}{d_G}\frac{n_H}{d_H}}x \\
&= \frac{n_Gd_H}{n_Gn_H + d_Gd_H}x \\
&= \frac{K(s-1)(s-2)}{K(s-1)(s-2) + (s+1)(s+2)}x \\
&= \frac{K(s-1)(s-2)}{K(s^2-3s+2) + (s^2+3s+2)}x \\
&= \frac{K(s-1)(s-2)}{(1+K)\left[s^2+\frac{3(1-K)}{1+K}s+\frac{2+K}{1+K}\right]}x
\end{align}
$\omega_n^2 = \frac{2+K}{1+K}$ and $2\zeta\omega_n = \frac{3(1-K)}{1+K}$. Therefore,
\begin{align}
\zeta &= \frac{3(1-K)}{2\omega_n} = \frac{3(1-K)}{2\sqrt{2+K}\sqrt{1+K}}
\end{align}
If you want $\zeta = \frac{1}{\sqrt{2}}$ then $K= \frac{12-\sqrt{109}}{7} $
Best Answer
This is probably polynomial long-division to two terms, treating the denominator with higher degree (since the Transfer function is proper). In particular, you can verify that the denominator poly $s^n + a_{M-1} s^{n-1} + \dots + a_0$ is equal to $$ (s^{n-M}+(a_{n-1}-b_{M-1})s^{n-M-1})(s^M + b_{M-1} s^{M-1} + \dots + b_0) + O(s^{n-2}) $$ so we can substitute that into our root locus equation to yield $$ 1 + K \frac{s^M + b_{M-1} s^{M-1} + \dots + b_0}{(s^{n-M}+(a_{n-1}-b_{M-1})s^{n-M-1})(s^M + b_{M-1} s^{M-1} + \dots + b_0) + O(s^{n-2})} $$ Dividing through by the numerator gives $$ 1+ K \frac{1}{(s^{n-M}+(a_{n-1}-b_{M-1})s^{n-M-1}) + O(s^{n-M-2})} $$ With that we eliminate the latter, lower order terms to arrive at the desired expression.