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 $x$?

HINT:
$$\displaystyle \frac{d\left(\frac{x^2-1}{x^2+1}\right)^3}{dx} =\frac{d\left(\frac{x^2-1}{x^2+1}\right)^3}{d\left(\frac{x^2-1}{x^2+1}\right)}\cdot\frac{d\left(\frac{x^2-1}{x^2+1}\right)}{dx}$$
Now, $$\frac{d\left(\frac{x^2-1}{x^2+1}\right)^3}{d\left(\frac{x^2-1}{x^2+1}\right)}=3\left(\frac{x^2-1}{x^2+1}\right)^2$$
and $$\frac{d\left(\frac{x^2-1}{x^2+1}\right)}{dx}=\frac{d\left(1-\frac2{x^2+1}\right)}{dx}=\frac{2\cdot2x}{(x^2+1)^2}$$
