Since $f(x-1)=\frac{\sin \pi(x-1)}{\sqrt{1+x-1+(x-1)^{2}}}=\frac{-\sin \pi x}{\sqrt{1-x+x^{2}}}=f(-x)$,
the graph of $f(x)$ is symmetric about the line $x=-\frac{1}{2}$.
When $x=-\frac{1}{2}$, $\sin \pi x$ reaches its minimum value $-1$, and $\frac{1}{\sqrt{1+x+x^{2}}}$ reaches its maximum value $\frac{2 \sqrt{3}}{3}$. Therefore, the minimum value of $f(x)$ is $f\left(-\frac{1}{2}\right)=-\frac{2 \sqrt{3}}{3}$.
6. The minimum value of the function defined on the real numbers
$f(x)=\frac{\sin \pi x}{\sqrt{1+x+x^{2}}}$ is $\qquad$ -