Find the vertical and horizontal asymptotes of the function $f(x)=\frac{3x^2-2x+1}{x^2-1}$.

Step 1 ：Step 1: To find the vertical asymptotes, we set the denominator to zero and solve for x. So, $x^2-1=0$.

Step 2 ：Step 2: Factoring the equation $x^2-1=0$ gives $(x-1)(x+1)=0$.

Step 3 ：Step 3: Setting each factor equal to zero gives the possible vertical asymptotes as $x=1$ and $x=-1$. However, we need to check if these values also make the numerator zero. If they do, they are not asymptotes. Substituting these values into the numerator shows that they do not make the numerator zero, so $x=1$ and $x=-1$ are indeed vertical asymptotes.

Step 4 ：Step 4: To find the horizontal asymptotes, we compare the degrees of the numerator and denominator. If the degrees are the same, the horizontal asymptote is the ratio of the leading coefficients. Here, the degrees of the numerator and denominator are both 2, so the horizontal asymptote is $y=\frac{3}{1}=3$.