Answer
$$\frac{{dy}}{{dx}} = \frac{{6\sqrt {y - 1} }}{{{x^{1/3}}\left( {1 - \sqrt {y - 1} } \right)}}$$
Work Step by Step
$$\eqalign{
& 2\sqrt {y - 1} = 9{x^{2/3}} + y \cr
& {\text{write }}\sqrt {y - 1{\text{ }}} {\text{ as }}{\left( {y - 1} \right)^{1/2}} \cr
& 2{\left( {y - 1} \right)^{1/2}} = 9{x^{2/3}} + y \cr
& {\text{take the derivative on both sides with respect to }}x \cr
& \frac{d}{{dx}}\left( {2{{\left( {y - 1} \right)}^{1/2}}} \right) = \frac{d}{{dx}}\left( {9{x^{2/3}}} \right) + \frac{d}{{dx}}\left( y \right) \cr
& {\text{solve the derivatives using the chain rule}} \cr
& 2\left( {\frac{1}{2}} \right){\left( {y - 1} \right)^{ - 1/2}}\frac{d}{{dx}}\left( {y - 1} \right) = 9\left( {\frac{2}{3}} \right){x^{ - 1/3}} + \frac{{dy}}{{dx}} \cr
& {\left( {y - 1} \right)^{ - 1/2}}\frac{{dy}}{{dx}} = 6{x^{ - 1/3}} + \frac{{dy}}{{dx}} \cr
& {\text{simplifying}} \cr
& \frac{1}{{\sqrt {y - 1} }}\frac{{dy}}{{dx}} = 6{x^{ - 1/3}} + \frac{{dy}}{{dx}} \cr
& {\text{combine terms and solve for }}\frac{{dy}}{{dx}} \cr
& \frac{1}{{\sqrt {y - 1} }}\frac{{dy}}{{dx}} - \frac{{dy}}{{dx}} = 6{x^{ - 1/3}} \cr
& \left( {\frac{1}{{\sqrt {y - 1} }} - 1} \right)\frac{{dy}}{{dx}} = 6{x^{ - 1/3}} \cr
& \left( {\frac{{1 - \sqrt {y - 1} }}{{\sqrt {y - 1} }}} \right)\frac{{dy}}{{dx}} = 6{x^{ - 1/3}} \cr
& \frac{{dy}}{{dx}} = \frac{{6\sqrt {y - 1} }}{{{x^{1/3}}\left( {1 - \sqrt {y - 1} } \right)}} \cr} $$
