Answer
$$\frac{{dy}}{{dx}} = - \frac{y}{x}$$
Work Step by Step
$$\eqalign{
& {x^2}{y^2} = 1 \cr
& {\text{differentiate both sides with respect to }}x \cr
& \frac{d}{{dx}}\left( {{x^2}{y^2}} \right) = \frac{d}{{dx}}\left( 1 \right) \cr
& {\text{use the product rule}} \cr
& {x^2}\frac{d}{{dx}}\left( {{y^2}} \right) + {y^2}\frac{d}{{dx}}\left( {{x^2}} \right) = \frac{d}{{dx}}\left( 1 \right) \cr
& {\text{Find derivatives}}{\text{, }} \cr
& {x^2}\left( {2y\frac{{dy}}{{dx}}} \right) + {y^2}\left( {2x} \right) = 0 \cr
& 2{x^2}y\frac{{dy}}{{dx}} + 2x{y^2} = 0 \cr
& \cr
& {\text{Solve for }}\frac{{dy}}{{dx}} \cr
& 2{x^2}y\frac{{dy}}{{dx}} = - 2x{y^2} \cr
& \frac{{dy}}{{dx}} = - \frac{{2x{y^2}}}{{2{x^2}y}} \cr
& \frac{{dy}}{{dx}} = - \frac{y}{x} \cr} $$
