I cannot see how to derive the following optimal portfolio weight $w^*$:
$$ w^* = \frac {(b - \frac{1}{2}) \rho \sigma_1\sigma_2 - b\sigma_2^2 +\sigma_1\sigma_2\sqrt{(b - \frac{1}{2})^2\rho^2 + b(1-b)}} {(1-b)\sigma_1^2 - b\sigma_2^2 + 2(b - \frac{1}{2})\rho\sigma_1\sigma_2} $$
From this system of equations:
$$ \begin{bmatrix} b \\ 1-b \\ \end{bmatrix} = \frac{1}{\sigma(x)} \cdot \begin{bmatrix} w^2\sigma_1^2 + \rho w(1-w) \sigma_1 \sigma_2 \\ (1-w)^2\sigma_2^2 + \rho w(1-w) \sigma_1 \sigma_2 \\ \end{bmatrix} $$
Where:
$$ \sigma(x) = \sqrt{w^2\sigma_1^2 + (1-w)^2\sigma_2^2 + 2 \rho w(1-w) \sigma_1\sigma_2} $$
The result is from a finance book called Introduction to Risk Parity and Budgeting (2013) by Thierry Roncalli. The derivation is omitted in the text and I can't seem to do it myself, and I cannot find a derivation for this particular case anywhere.
Any advice on how to solve the system would be a great help.
You can solve this by reducing it to the form of a single quadratic equation in $w$.
Eliminate the expression $\sigma(x)$ from the equation relating the column vectors by looking at the quotient of top and bottom column entries. The quotient you obtain is
$\frac{1-b}{b}=\frac {(1- w)^2 \sigma_2^2+ \rho \sigma_1 \sigma_2 w(1-w)}{w^2 \sigma_1^2 + w(1-w)\rho \sigma_1 \sigma_2 }$
Then clear of denominators by cross-multiplying to obtain a quadratic equation that $w$ must satisfy. After expanding out everything in powers of $w$ you get a quadratic that has the form $Aw^2 + B w+ C=0$.
Explicitly the individual terms in the quadratic have this structure:
$$Aw^2=w^2 [( 2 b-1) \rho \sigma _2 \sigma _1 +(1-b) \sigma _1^2-b \sigma_2^2]$$
$$Bw= w \left(-2 b \rho \sigma _1 \sigma _2+2 b \sigma _2^2+\rho \sigma _1 \sigma _2\right)= w ((1-2 b )\rho \sigma _1 \sigma _2+2 b \sigma _2^2)$$
$$C= -b \sigma _2^2$$
The solutions to the quadratic are $w = \frac{-B/2 \pm \sqrt{ (B/2)^2- AC}}{A}$.
Note that the denominator term $A$ matches that in the textbook. Perhaps these pointers will allow you to wrap up the algebra to check also that the numerators agree.