\[f(x) = \frac{x^{3} + 1}{3}\]
\[1)\ Пересечение\ с\ Ox:\]
\[\frac{x^{3} + 1}{3} = 0\]
\[x^{3} + 1 = 0\]
\[x^{3} = - 1\]
\[x = - 1.\]
\[2)\ f^{'}(x) = \frac{3x^{2} + 0}{3} = \frac{3x^{2}}{3} = x^{2};\]
\[f^{'}( - 1) = ( - 1)^{2} = 1;\]
\[f( - 1) = \frac{( - 1)^{3} + 1}{3} = \frac{0}{3} = 0;\]
\[y = 0 + 1 \bullet (x + 1) = x + 1.\]
\[Ответ:\ \ y = x + 1.\]