Answer
$ y'=\frac{9}{4}x^{\frac{1}{2}}y^{\frac{5}{3}}$
Work Step by Step
Differentiating implicitly with respect to $ x $, we have
$-\frac{2}{3}y^{-\frac{5}{3}}y'+\frac{3}{2}x^{\frac{1}{2}}=0$
$-\frac{2}{3}y^{-\frac{5}{3}}y'=-\frac{3}{2}x^{\frac{1}{2}}$
$\implies y'=-\frac{3}{2}x^{\frac{1}{2}}\times-\frac{3}{2}y^{\frac{5}{3}}$
$=\frac{9}{4}x^{\frac{1}{2}}y^{\frac{5}{3}}$
