6. The solution to the equation $2^{x^{2}-2 x}+3^{x^{2}-2 x}+x^{2}-2 x+\frac{1}{6}=0$ is
6. $x=1$.

Let $f(t)=2^{t}+3^{t}+t$, then the original equation becomes $f\left(x^{2}-2 x\right)=-\frac{1}{6}=f(-1)$. Since $f(t)$ is monotonically increasing, we have $x^{2}-2 x=-1$, which gives $x=1$.