12．已知函数f(x)={x2+1x(x<0)x2−1x(x>0)f(x)=\left\{{\left.\begin{array}{l}{\frac{{x^2}+1}{x}({x<0})}\\ {\frac{{x^2}-1}{x}({x>0})}\end{array}\right.}\right.f(x)={xx2+1(x<0)xx2−1(x>0)，则方程f2(x)−f(x)−6=0f^{2}(x)-f(x)-6=0f2(x)−f(x)−6=0的实根个数为((( )))