Answer
$$\frac{{dy}}{{dx}} = \frac{{ - 4{x^{1/3}}{y^{4/3}} + {x^{ - 2/3}}}}{{4{x^{4/3}}{y^{1/3}} - 18{y^5}}}$$
Work Step by Step
$$\eqalign{
& {\left( {xy} \right)^{4/3}} + {x^{1/3}} = {y^6} + 1 \cr
& {\text{use the property }}{\left( {ab} \right)^n}{a^n}{b^n} \cr
& {x^{4/3}}{y^{4/3}} + {x^{1/3}} = {y^6} + 1 \cr
& {\text{take the derivative on both sides with respect to }}x \cr
& \frac{d}{{dx}}\left( {{x^{4/3}}{y^{4/3}} + {x^{1/3}}} \right) = \frac{d}{{dx}}\left( {{y^6} + 1} \right) \cr
& {\text{use sum rule and product rule for differentiation as follows}} \cr
& \frac{d}{{dx}}\left( {{x^{4/3}}{y^{4/3}}} \right) + \frac{d}{{dx}}\left( {{x^{1/3}}} \right) = \frac{d}{{dx}}\left( {{y^6}} \right) + \frac{d}{{dx}}\left( 1 \right) \cr
& {x^{4/3}}\frac{d}{{dx}}\left( {{y^{4/3}}} \right) + {y^{4/3}}\frac{d}{{dx}}\left( {{x^{4/3}}} \right) + \frac{d}{{dx}}\left( {{x^{1/3}}} \right) = \frac{d}{{dx}}\left( {{y^6}} \right) + \frac{d}{{dx}}\left( 1 \right) \cr
& {\text{solve the derivatives using the power rule and the chain rule}} \cr
& {x^{4/3}}\left( {\frac{4}{3}{y^{1/3}}} \right)\frac{{dy}}{{dx}} + {y^{4/3}}\left( {\frac{4}{3}{x^{1/3}}} \right) + \frac{1}{3}{x^{ - 2/3}} = 6{y^5}\frac{{dy}}{{dx}} + 0 \cr
& \frac{4}{3}{x^{4/3}}{y^{1/3}}\frac{{dy}}{{dx}} + \frac{4}{3}{x^{1/3}}{y^{4/3}} + \frac{1}{3}{x^{ - 2/3}} = 6{y^5}\frac{{dy}}{{dx}} \cr
& {\text{combine terms and solve for }}\frac{{dy}}{{dx}} \cr
& \frac{4}{3}{x^{4/3}}{y^{1/3}}\frac{{dy}}{{dx}} - 6{y^5}\frac{{dy}}{{dx}} = - \frac{4}{3}{x^{1/3}}{y^{4/3}} + \frac{1}{3}{x^{ - 2/3}} \cr
& \left( {\frac{4}{3}{x^{4/3}}{y^{1/3}} - 6{y^5}} \right)\frac{{dy}}{{dx}} = - \frac{4}{3}{x^{1/3}}{y^{4/3}} + \frac{1}{3}{x^{ - 2/3}} \cr
& {\text{multiply both sides by 3}} \cr
& \left( {4{x^{4/3}}{y^{1/3}} - 18{y^5}} \right)\frac{{dy}}{{dx}} = - 4{x^{1/3}}{y^{4/3}} + {x^{ - 2/3}} \cr
& \frac{{dy}}{{dx}} = \frac{{ - 4{x^{1/3}}{y^{4/3}} + {x^{ - 2/3}}}}{{4{x^{4/3}}{y^{1/3}} - 18{y^5}}} \cr} $$
