#### Answer

$ f'(0)=0$,
Equation of the tangent line is:
$ y=1$

#### Work Step by Step

Recall that $ f'(a)=\lim\limits_{h \to 0}\frac{f(a+h)-f(a)}{h}$
Given that $ a=0$ and $ f(x)=\frac{1}{x^{2}+1}$. Therefore, $ f'(0)=\lim\limits_{h \to 0}\frac{f(0+h)-f(0)}{h}=\lim\limits_{h \to 0}\frac{\frac{1}{h^{2}+1}-1}{h}$
$=\lim\limits_{h \to 0}\frac{1-h^{2}-1}{h(h^{2}+1)}=\lim\limits_{h \to 0}\frac{-h}{h^{2}+1}=0$
Equation of the tangent line is of the form
$ y-f(a)=f'(a)(x-a)$
Knowing that $ f(a)=f(0)=1, f'(a)=f'(0)=0$ and $ a=0$, we obtain the equation of the tangent line through $ a $ as below.
$ y-1=0(x-0)$
Or $ y=1$