Answer
Local maximum.
Work Step by Step
Step 1. Given $f(x)=\frac{ax+b}{x^2-1}$, we have $f'(x)=\frac{a(x^2-1)-2x(ax+b)}{(x^2-1)^2}=\frac{-ax^2-2bx-a}{(x^2-1)^2}$
Step 2. Since $x=3$ is a critical point, we have $f'(3)=0$, which leads to $a(3)^2+2b(3)+a=0$ or $5a+3b=0$
Step 3. As $f(3)=1$, we have $\frac{a(3)+b}{3^2-1}=1$ or $3a+b=8$
Step 4. Solve the system of equations from steps 2 and 3 to get $a=6, b=-10$
Step 5. With $f'(x)=-\frac{6x^2-20x+6}{(x^2-1)^2}=-\frac{2(3x-1)(x-3)}{(x^2-1)^2}$; testing sign changes of $f'(x)$ across $x=3$, we have $(1)..(+)..(3)..(-)..$ indicating a concave down region with a local maximum.
Step 6. We conclude that $f(3)=1$ is a local maximum of the function as shown in the figure.
