Answer
$$\frac{1}{{15{y^2} + 1}}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = 5{x^3} + x - 7 \cr
& {\text{By the formula }}\left( {\text{3}} \right){\text{: }}\frac{{dy}}{{dx}}{\text{ = }}\frac{1}{{dx/dy}} \cr
& {\text{Differentiate }}f\left( y \right) = 5{y^3} + y - 7{\text{ with respect to }}y \cr
& y = \left[ {5{x^3} + x - 7} \right] \cr
& 1 = 15{x^2}\frac{{dx}}{{dy}} + \frac{{dx}}{{dy}} - 0 \cr
& {\text{Solve for }}\frac{{dx}}{{dy}} \cr
& \left( {15{x^2} + 1} \right)\frac{{dx}}{{dy}} = 1 \cr
& \frac{{dx}}{{dy}} = \frac{1}{{15{x^2} + 1}} \cr
& f\left( y \right) = x,{\text{ then }} \cr
& \frac{{dy}}{{dx}} = \frac{1}{{15{y^2} + 1}} \cr
& \cr
& {\text{Differentiating implicitly}} \cr
& y = 5{x^3} + x - 7 \cr
& {\text{Let }}x = f\left( y \right) \cr
& x = 5{y^3} + y - 7 \cr
& \frac{d}{{dx}}\left[ x \right] = \frac{d}{{dx}}\left[ {5{y^3} + y - 7} \right] \cr
& 1 = 15{y^2}\frac{{dy}}{{dx}} + \frac{{dy}}{{dx}} - 0 \cr
& {\text{Solve for }}\frac{{dy}}{{dx}} \cr
& 1 = \left( {15{y^2} + 1} \right)\frac{{dy}}{{dx}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{{15{y^2} + 1}} \cr} $$
