Answer
$$\frac{1}{{5 - 2\cos 2y}}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = 5x - \sin 2x,{\text{ }} - \frac{\pi }{4} < x < \frac{\pi }{4} \cr
& {\text{By the formula }}\left( {\text{3}} \right){\text{: }}\frac{{dy}}{{dx}}{\text{ = }}\frac{1}{{dx/dy}} \cr
& {\text{Differentiate }}f\left( y \right) = 5{y^3} + y - 7{\text{ with respect to }}y \cr
& y = \left[ {5x - \sin 2x} \right] \cr
& 1 = 5\frac{{dx}}{{dy}} - 2\cos 2x\frac{{dx}}{{dy}} \cr
& {\text{Solve for }}\frac{{dx}}{{dy}} \cr
& \left( {5 - 2\cos 2x} \right)\frac{{dx}}{{dy}} = 1 \cr
& \frac{{dx}}{{dy}} = \frac{1}{{5 - 2\cos 2x}} \cr
& f\left( y \right) = x,{\text{ then }} \cr
& \frac{{dy}}{{dx}} = \frac{1}{{5 - 2\cos 2y}} \cr
& \cr
& {\text{Differentiating implicitly}} \cr
& y = 5x - \sin 2x \cr
& {\text{Let }}x = f\left( y \right) \cr
& x = 5y - \sin 2y \cr
& \frac{d}{{dx}}\left[ x \right] = \frac{d}{{dx}}\left[ {5y - \sin 2y} \right] \cr
& 1 = 5\frac{{dy}}{{dx}} - 2\cos 2y\frac{{dy}}{{dx}} \cr
& {\text{Solve for }}\frac{{dy}}{{dx}} \cr
& 1 = \left( {5 - 2\cos 2y} \right)\frac{{dy}}{{dx}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{{5 - 2\cos 2y}} \cr} $$
