<s>[INST] Find points on curve $y = (x^2 - 1)^3 / (x^2 + 1)^3$ when slope of tangent line is zero? As soon as a get to the quotient rule I can't simplify to solve for $x$.
This is what I have tried:
Since the derivative much be equal to zero:
$$0 = [6x(x^2 -1)^2 (x^2+1)^3] - [6x(x^2 + 1)^2(x^2-1)^3]$$
How do I simplify further and solve for $x$?
 [/INST] Note that there is a common term $6x$, of $(x^2 - 1)^2$ and a common term of $(x^2 + 1)^2$, so your equation can be rewritten as
$$0 = 6x (x^2 - 1)^2 (x^2 + 1)^2 \Big((x^2 + 1) - (x^2 - 1)\Big) = 12x (x^2 - 1)^2(x^2 + 1)^2$$
Now set each term to $0$ individually.
</s>