Continuing @Paola Ramírez 's approach from inspiration problem:

$(a+b)^2=a^2+b^2+2ab=4ab+2ab=6ab$

$(a-b)^2=a^2+b^2-2ab=4ab-2ab=2ab$

Then, $\dfrac{(a+b)^2}{(a-b)^2}=\dfrac{6ab}{2ab} \Rightarrow \dfrac{a+b}{a-b}=\sqrt{3}$

The above is Paola's solution and we continue with it by applying componendo and dividendo like this:

$\dfrac{a+b}{a-b}=\dfrac{\sqrt{3}}{1} \implies \dfrac{a+b+a-b}{a+b-a+b}=\dfrac{\sqrt{3}+1}{\sqrt{3}-1}$

$\implies \dfrac{a}{b}=\dfrac{(\sqrt{3}+1)^2}{(\sqrt{3}-1)(\sqrt{3}+1)} = \dfrac{3+1+2\sqrt{3}}{3-1}=\dfrac{2(2+\sqrt{3})}{2}= 2+\sqrt{3}$

Since $\dfrac{1}{2+\sqrt{3}}=2-\sqrt{3} \implies \dfrac{b}{a}=2-\sqrt{3} \approx \boxed{0.268}$

Alternate Method:

$a^2+b^2=4ab$ and dividing both sides by $ab$ we have $\dfrac{a}{b}+\dfrac{b}{a}=4$ then let $x=\dfrac{b}{a}$ and we have $x+\dfrac{1}{x}=4 \implies x^2-4x+1=0$ and solving the quadratic gives $x=2\pm \sqrt{3}$ but since $b<a$ we have $\dfrac{b}{a}<1$ so we choose $2-\sqrt{3} \approx \boxed{0.268}$ since its less than $1$ .

This is a quadratic equation we can write $a^{2}+b^{2}-4ab=0$ lets solve for b: $b= \frac{4a +/- \sqrt{16a^{2}-4a^{2}}}{2}= \frac{4a+/- 2\sqrt{3}a}{2}= 2a+/-\sqrt{3}a$ Since $a>b => b=a(2- \sqrt{3}) \frac{b}{a}=\frac{a(2-\sqrt{3})}{a}$ Which leaves us with $2-\sqrt{3}=\boxed{0,268}$

As I skimmed through your solutions I realized mine is quite different. It goes like this:

1. $a^2+b^2=4ab$

2. $a^2-4ab+4b^2=3b^2$

3. $(a-2b)^2=3b^2)$

4. $a-2b=b\sqrt{3}$

5. $a=b(\sqrt{3}+2)$

Then, we can substitute $a$ for $b(\sqrt{3}+2)$ in $\frac{b}{a}$ leaving us with:

$\dfrac{b}{b(\sqrt{3}+2)}=\dfrac{1}{\sqrt{3}+2}\approx 0.268$

From the given information we see that

$(a+b)^2 = 6ab, (a-b)^2 = 2ab$

$\Rightarrow$ $\Large \frac{a+b}{a-b}$ = $\sqrt{3}$

Now, dividing numerator and denominator by a, we get,

$\Rightarrow \huge \frac{\frac{a+b}{a}}{\frac{a-b}{a}}$ = $\sqrt{3}$

$\Rightarrow \huge \frac{1 + \frac{b}{a}}{1 - \frac{b}{a}}$ = $\sqrt{3}$

$\Rightarrow \large 1 + \frac{b}{a}$ = $\sqrt{3} - \sqrt{3} \large \frac{b}{a}$

$\Rightarrow \large (1+\sqrt{3})\large \frac{b}{a}$ = $\sqrt{3}-1$

$\Rightarrow \Large \frac{b}{a}$ = $\Large \frac{\sqrt{3}-1}{\sqrt{3}+1}$ = $2 - \sqrt{3}$ $\approx 0.268$

a^2-4ab+b^2=0 =>> 3a^2-3a^2+a^2-4ab+b^2=0 =>> 4a^2-4ab+b^2=3a^2 =>>> (2a-b)^2=3a^2 =>>2a-b=a3^1/2 =>>>2a-a3^1/2=b =>> a(2-3^1/2)=b =>>> b/a =>>> a(2-3^1/2)/a=(2-3^1/2)

$1+(\frac b a )^2=4*\frac b a.\ \ \ Let\ \frac b a=X.\\ We\ know\ (m \pm n)^2=m^2 + n^2 \pm 2mn.\\ \implies\ (1+X)^2=6X,\\ \ \ and\ \ \ (1-X)^2=2X.\\ \therefore\ 1+X=\sqrt{6X}\ \ \ and\ \ \ \ \ 1-X=\sqrt{2X} \\ Adding\ the\ two\ equations,\\ \therefore\ 2=\sqrt{2X}*(\sqrt3+1) \\ \implies\ X=\dfrac 2{4+2\sqrt3}=2-\sqrt3=\large \ \ \ \ \color{#D61F06}{0.2679}$

Divide by a ^2 [b/a]^2-4[b/a] +1=0 so b/a 2+√3. or 2 -√3. b/a <1 so b/a = 2-√3 = 0.268