Calc pls help critical points already have 2/3

$f(x) = \dfrac{3x^2 + 3x + 6}{x^2 + x - 56} = 3 - \dfrac{58}5 \left(\dfrac{1}{x + 8} - \dfrac1{x - 7}\right)$

Differentiating, $f'(x) = \dfrac{58}5\left(\dfrac1{(x + 8)^2} - \dfrac1{(x - 7)^2}\right)$

When $f'(x) = 0$,

$(x - 7)^2 - (x + 8)^2 = 0\\ 2x + 1 = 0\\ x = -\dfrac12$

Also, $f\left(-\dfrac12\right) = -\dfrac7{75}$, so the critical point is $\left(-\dfrac12, -\dfrac7{75}\right)$, and there is only 1 critical point, not 3.