Answer
$\frac{{dy}}{{dx}} = \frac{{{{\sec }^2}\left( {x - y} \right) - 2{y^3}}}{{{{\sec }^2}\left( {x - y} \right) + 6x{y^2}}}$
Work Step by Step
$$\eqalign{
& \tan \left( {x - y} \right) = 2x{y^3} + 1 \cr
& {\text{Differentiating both sides of the equation with respect to }}x \cr
& \frac{d}{{dx}}\left[ {\tan \left( {x - y} \right)} \right] = \underbrace {\frac{d}{{dx}}\left[ {2x{y^3}} \right]}_{{\text{use product rule}}} + \frac{d}{{dx}}\left[ 1 \right] \cr
& \frac{d}{{dx}}\left[ {\tan \left( {x - y} \right)} \right] = 2x\frac{d}{{dx}}\left[ {{y^3}} \right] + 2{y^3}\frac{d}{{dx}}\left[ x \right] + \frac{d}{{dx}}\left[ 1 \right] \cr
& y{\text{ is a function of }}x,{\text{ using the chain rule we have}} \cr
& {\sec ^2}\left( {x - y} \right)\frac{d}{{dx}}\left[ {x - y} \right] = 2x\left( {3{y^2}} \right)\frac{{dy}}{{dx}} + 2{y^3}\left( 1 \right) + 0 \cr
& {\sec ^2}\left( {x - y} \right)\left( {1 - \frac{{dy}}{{dx}}} \right) = 6x{y^2}\frac{{dy}}{{dx}} + 2{y^3} \cr
& {\sec ^2}\left( {x - y} \right) - {\sec ^2}\left( {x - y} \right)\frac{{dy}}{{dx}} = 6x{y^2}\frac{{dy}}{{dx}} + 2{y^3} \cr
& {\text{Collect the terms that contains }}\frac{{dy}}{{dx}} \cr
& - {\sec ^2}\left( {x - y} \right)\frac{{dy}}{{dx}} - 6x{y^2}\frac{{dy}}{{dx}} = 2{y^3} - {\sec ^2}\left( {x - y} \right) \cr
& {\sec ^2}\left( {x - y} \right)\frac{{dy}}{{dx}} + 6x{y^2}\frac{{dy}}{{dx}} = {\sec ^2}\left( {x - y} \right) - 2{y^3} \cr
& {\text{Solve this equation for }}\frac{{dy}}{{dx}} \cr
& \left[ {{{\sec }^2}\left( {x - y} \right) + 6x{y^2}} \right]\frac{{dy}}{{dx}} = {\sec ^2}\left( {x - y} \right) - 2{y^3} \cr
& \frac{{dy}}{{dx}} = \frac{{{{\sec }^2}\left( {x - y} \right) - 2{y^3}}}{{{{\sec }^2}\left( {x - y} \right) + 6x{y^2}}} \cr} $$
