Answer
$ f'(0)=1$
Equation of the tangent line:
$ y=x $
Work Step by Step
Recall that $ f'(a)=\lim\limits_{h \to 0}\frac{f(a+h)-f(a)}{h}$
Given that $ a=0$. Therefore, $ f'(0)=\lim\limits_{h \to 0}\frac{f(0+h)-f(0)}{h}=\lim\limits_{h \to 0}\frac{(h^{3}+h)-(0^{3}+0)}{h}$
$=\lim\limits_{h \to 0}\frac{h^{3}+h}{h}=\lim\limits_{h \to 0}(h^{2}+1)=1$
Equation of the tangent line is of the form
$ y-f(a)=f'(a)(x-a)$
Knowing that $ f(a)=f(0)=0, f'(a)=f'(0)=1$ and $ a=0$, we obtain the equation of the tangent line through $ a $ as below.
$ y-0=1(x-0)$
Or $ y=x $
