Answer
$$\frac{{dy}}{{dx}} = \frac{{x + 1}}{y}{\text{ and }}\frac{{{d^2}y}}{{d{x^2}}} = \frac{{{y^2} - {{\left( {x + 1} \right)}^2}}}{{{y^3}}}$$
Work Step by Step
$$\eqalign{
& {y^2} = {x^2} + 2x \cr
& \cr
& {\text{differentiate both sides with respect to }}x \cr
& \frac{d}{{dx}}\left( {{y^2}} \right) = \frac{d}{{dx}}\left( {{x^2}} \right) + \frac{d}{{dx}}\left( {2x} \right) \cr
& {\text{find derivatives}} \cr
& 2y\frac{{dy}}{{dx}} = 2x + 2 \cr
& {\text{solve for }}\frac{{dy}}{{dx}} \cr
& \frac{{dy}}{{dx}} = \frac{{2x + 2}}{{2y}} \cr
& \frac{{dy}}{{dx}} = \frac{{x + 1}}{y} \cr
& \cr
& {\text{find the second derivative}} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = \frac{d}{{dx}}\left( {\frac{{x + 1}}{y}} \right) \cr
& {\text{use the quotient rule}} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = \frac{{y\frac{d}{{dx}}\left( {x + 1} \right) - \left( {x + 1} \right)\frac{d}{{dx}}\left( y \right)}}{{{y^2}}} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = \frac{{y - \left( {x + 1} \right)\frac{{dy}}{{dx}}}}{{{y^2}}} \cr
& {\text{substitute }}\frac{{dy}}{{dx}} = \frac{{x + 1}}{y}{\text{ and simplify}} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = \frac{{y - \left( {x + 1} \right)\left( {\frac{{x + 1}}{y}} \right)}}{{{y^2}}} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = \frac{{{y^2} - {{\left( {x + 1} \right)}^2}}}{{{y^3}}} \cr} $$
