conjecture：
Let $ O, $ $ F_1, $ $ F_2 $ be the circumcenter, 1st Fermat point, 2nd Fermat point of $ \triangle ABC, $ respectively. Prove that $$ \text{Power of } F_1 \text{ WRT } \odot (O) \text{ + Power of } F_2 \text{ WRT } \odot (O) \text{ = } -{F_1F_2}^2. $$(the power of WRT see: Power WRT)
I have use GeoGebra test this conjecture are right:see this following 
But How to prove this
I think I have a proof, but I'm unsure if my brain counts as a credible and/or official source :P
Let $\Gamma$ be the circumcircle. Let $G,H,N$ be the centroid, orthocenter and 9-point center. Let $M,X$ be the midpoints of $\overline{HG},\overline{F_1F_2}$. The following facts are well-known:
Thus from the first fact $$MG^2=MF_1\cdot MF_2=MX^2-\frac{1}{4}F_1F_2^2\implies\boxed{MX^2=\frac{1}{4}\left(GH^2+F_1F_2^2\right)}.$$ But as $G,H$ are the internal and external similitude centers of $\Gamma$ and the 9-point circle, so $\omega$ is coaxal with these circles. Thus $$\frac{\operatorname{Pow}_{\Gamma}(X)}{\operatorname{Pow}_{\omega}(X)}=\frac{XO^2-R^2}{XM^2-\frac{1}{4}GH^2}=\frac{NO}{NM}=-\frac{NO}{NG}=-3,$$ so $\boxed{XO^2-R^2=-\frac{3}{4}F_1F_2^2}$. By Apollonius' median theorem in $\triangle OF_1F_2$, $$\boxed{XO^2=\frac{1}{2}\left(OF_1^2+OF_2^2\right)-\frac{1}{4}F_1F_2^2},$$ so putting the two together, $$\operatorname{Pow}_{\Gamma}(F_1)+\operatorname{Pow}_{\Gamma}(F_2)=OF_1^2+OF_2^2-2R^2=2XO^2+\frac{1}{2}F_1F_2^2-2R^2=-F_1F_2^2$$
As mentioned by OP, the question already has a short ---and quite sophisticated--- answer on the Art of Problem Solving site, by AoPS user Luis González. Since OP has requested an alternative approach, I'll outline mine ... which ---fair warning!--- is Tedium incarnate. (I said from the beginning that the proof was "ugly".)
I'll start with the characterization of the Fermat points as points of concurrency of lines associated with the two Napoleon Triangle-related configurations:
Given $\triangle ABC$, erect "outer" equilateral triangles $\triangle D_{+}BC$, $\triangle E_{+}CA$, $\triangle F_{+}AB$ that do not overlap $\triangle ABC$; and erect "inner" equilateral triangles $\triangle D_{-}BC$, $\triangle E_{-}CA$, $\triangle F_{-}AB$ that do overlap $\triangle ABC$. Then $\overleftrightarrow{AD_{+}}$, $\overleftrightarrow{BE_{+}}$, $\overleftrightarrow{CF_{+}}$ concur at the first Fermat point (here, $K_{+}$), while $\overleftrightarrow{AD_{-}}$, $\overleftrightarrow{BE_{-}}$, $\overleftrightarrow{CF_{-}}$ concur at the second Fermat point ($K_{-}$).
Coordinatizing, with the circumcenter of $\triangle ABC$ at the origin ($O$), we can take $$A = r(1,0) \qquad B = r(\cos 2C, \sin 2C) \qquad C = r ( \cos 2B, - \sin 2B) $$ where $r$ is the circumradius. Note that, for instance, $O$, $E_{+}$, $E_{-}$ are collinear with the midpoint, $M$, of $\overline{AC}$; in particular, the reader can verify that $$E_{\pm} = \frac{\cos B \pm \sqrt{3} \sin B}{\cos B} \;M = 2 r \cos\left(B\mp \frac{\pi}{3}\right)\;(\cos B, -\sin B)$$ Likewise, $$F_{\pm} = 2 r \cos\left(C\mp \frac{\pi}{3}\right)\;(\cos C, \sin C)$$
From here, one can exploit a computer algebra system to determine equations for lines $\overleftrightarrow{BE_{\pm}}$ and $\overleftrightarrow{CF_{\pm}}$, and to generate coordinates of their intersections (the Fermat points, $K_{\pm}$). With a bit of fiddling, one can express the coordinates in this symmetric form:
$$K_{\pm} = \frac{ \left(\quad\begin{align} A \; \sin B \; ( 4 \sin B \sin C \pm \sqrt{3} \sin A ) \\ + \; B \; \sin B \; ( 4 \sin C \sin A \pm \sqrt{3} \sin B ) \\ + \; C \; \sin C \; ( 4 \sin A \sin B \pm \sqrt{3} \sin C ) \end{align} \quad\right) }{ 6 \sin A \sin B \sin C \pm \sqrt{3} ( \sin^2A + \sin^2 B + \sin^2C ) } \tag{1}$$
Alternatively, with $a = |\overline{BC}|$, etc, we have a somewhat-more-compact representation: $$K_{\pm} = \frac{ 2 a b c ( A + B + C ) \pm r \sqrt{3} \left( a^2 A + b^2 B + c^2 C \right)}{ 3 a b c \pm r \sqrt{3} \left(\;a^2 + b^2 + c^2 \right)} \tag{2}$$
We could even write $$K_{\pm} = \pm \;\frac{ A ( 2 T \pm T_a ) + B ( 2 T \pm T_b ) + C ( 2 T \pm T_c ) }{6 T_{\pm}} \tag{3}$$ where $T := |\triangle ABC|$, and $T_a$, $T_b$, $T_c$ are the areas of the equilateral triangles, and $T_{\pm}$ is the area of the corresponding inner-or-outer Napoleon Triangle. (Note: $T = T_{+} - T_{-}$.)
I'll also note that $$\begin{align} &6 \sin A \sin B \sin C \pm \sqrt{3} \left( \sin^2A + \sin^2B + \sin^2C \right) \\ =\;&2 \sqrt{3} \left(\;\sin A \sin\left(A \pm \frac{\pi}{3}\right) + \sin B \sin\left(B \pm \frac{\pi}{3}\right) + \sin C \sin\left(C \pm \frac{\pi}{3}\right)\;\right) \end{align} \tag{4}$$ which makes me think that the numerator has a clever form involving $\pm \frac{\pi}{3}$, although I never found anything completely satisfying.
I've been stalling, because the remainder of this "proof" of the relation
$$\operatorname{pow}(K_{+}) + \operatorname{pow}(K_{-}) = -|\overline{K_{+}K_{-}}|^2 \tag{$\star$}$$
amounts to letting the computer algebra system verify this equivalent form (leveraging the fact that the center of the circle in question lies at the origin): $$|K_{+}|^2 + |K_{-}|^2 - (K_{+}\cdot K_{-}) - r^2 = 0 \tag{$\star\star$}$$ where we invoke the relations $$|A|^2 = |B|^2 = |C|^2 = r^2, \quad B\cdot C = r^2 \cos 2A, \quad C\cdot A = r^2 \cos 2B, \quad A\cdot B = r^2 \cos 2C$$ $$A + B + C = \pi$$
... making for an anti-climactic, and wholly-unsatisfying, QED. $\square$
The reason I have so many alternative representations of $K_{\pm}$ is that I was looking for a form that makes relation $(\star\star)$ "obvious", or something that would provide a helpful larger context under which $(\star)$ holds. I have failed in these quests so far. Unless and until profound inspiration strikes, the above is all I have to offer.
