Answer
$$\frac{{dy}}{{dx}} = - \frac{{{y^3}}}{2}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{1}{{{x^2}}} \cr
& {\text{By the formula }}\left( {\text{3}} \right){\text{: }}\frac{{dy}}{{dx}}{\text{ = }}\frac{1}{{dx/dy}} \cr
& {\text{Differentiate }}f\left( x \right) = 5{x^3} + x - 7{\text{ with respect to }}y \cr
& f\left( x \right) = \frac{1}{{{x^2}}} \cr
& y = \frac{1}{{{x^2}}} \cr
& \frac{d}{{dy}}\left[ y \right] = \frac{d}{{dy}}\left[ {\frac{1}{{{x^2}}}} \right] \cr
& 1 = - \frac{2}{{{x^3}}}\frac{{dx}}{{dy}} \cr
& {\text{Solve for }}\frac{{dx}}{{dy}} \cr
& 1 = - \frac{2}{{{x^3}}}\frac{{dx}}{{dy}} \cr
& \frac{{dx}}{{dy}} = - \frac{{{x^3}}}{2} \cr
& f\left( y \right) = x,{\text{ then }} \cr
& \frac{{dy}}{{dx}} = - \frac{{{y^3}}}{2} \cr
& \cr
& {\text{Differentiating implicitly}} \cr
& y = \frac{1}{{{x^2}}} \cr
& {\text{Let }}x = f\left( y \right) \cr
& x = \frac{1}{{{y^2}}} \cr
& \frac{d}{{dx}}\left[ x \right] = \frac{d}{{dx}}\left[ {\frac{1}{{{y^2}}}} \right] \cr
& 1 = - \frac{2}{{{y^3}}}\frac{{dy}}{{dx}} \cr
& {\text{Solve for }}\frac{{dy}}{{dx}} \cr
& \frac{{dy}}{{dx}} = - \frac{{{y^3}}}{2} \cr} $$
