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