In the triangle $ABC$, if $a=\frac{2(b^2-c^2)}{-b+\sqrt{b^2+4c^2}}$, prove that $3\cdot\widehat{C}=2\cdot\widehat{B}$. Just like in the title, I have to prove that if in a triangle $ABC$
$$a=\frac{2(b^2-c^2)}{-b+\sqrt{b^2+4c^2}}$$
holds, then $3\cdot\widehat{C}=2\cdot\widehat{B}$.

The denominator of the big fraction looks like the positive solution
of a quadratic equation, specifically  $x^2 + bx - c^2 = 0$, so the claim appears to be that if $ x = \frac{b^2-c^2}{a}$ is a solution of that equation, then $\frac{\widehat{C}}{\widehat{B}}=\frac{2}{3}$.
I have not gone beyond this examination of the problem itself, but
perhaps this will give you some ideas to try.

Assuming $\widehat{ C}=2\theta, \widehat{B}=3\theta,\widehat{A}=\pi-5\theta$ and $R=1$ (the circumradius) we have:
$$ a = 2\sin(5\theta),\quad b=2\sin(3\theta),\quad c=2\sin(2\theta) $$
and $\frac{b^2-c^2}{a}=2\sin\theta$. On the other hand,
$$ \sin^2(\theta) + \sin(3\theta)\sin(\theta)-\sin^2(2\theta)=0.$$
