## Calculus (3rd Edition)

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