Answer
$$y = x$$
Work Step by Step
$$\eqalign{
& y = 3{x^3} + \sin x,{\text{ point }}\left( {0,0} \right) \cr
& {\text{differentiate }}y \cr
& y'\left( x \right) = \frac{d}{{dx}}\left[ {3{x^3} + \sin x} \right] \cr
& y'\left( x \right) = 9{x^2} + \cos x \cr
& {\text{find the slope at the point }}\left( {0,0} \right) \cr
& \,\,\,m = y'\left( 0 \right) \cr
& \,\,\,m = 9{\left( 0 \right)^2} + \cos \left( 0 \right) \cr
& \,\,\,m = 1 \cr
& {\text{find the equation of the tangent line at the point }}\left( {0,0} \right) \cr
& \,\,\,\,\,y - {y_1} = m\left( {x - {x_1}} \right) \cr
& \,\,\,\,\,y - 0 = 1\left( {x - 0} \right) \cr
& {\text{simplify}} \cr
& \,\,\,\,\,y = x \cr} $$
