Answer
$y=-1$
Work Step by Step
The equation of the tangent line to $f$ at $(0,-1)$ is:
$$y=-1+f'(0)(x-0)$$
$$f'(x)=\frac{2x(x^{2}+1)-(x^{2}-1)2x}{(x^{2}+1)^{2}}$$
$$f'(0)=\frac{2\cdot (0)(0^{2}+1)-(0^{2}-1)2(0)}{(0^{2}+1)^{2}}=0$$
so:
$$y=-1+f'(0)(x-0) \to y=-1+0(x-0) \to y=-1$$
so the tangent line is:
$$y=-1$$
