Answer
$$y'(x)= -\frac{3 y^{\frac{1}{3}}}{4 \sqrt{x} \cdot\left(1+6 y^{\frac{1}{3}}\right)} $$
Work Step by Step
Given $$x^{1 / 2}+y^{2 / 3}=-4 y$$
Differentiate with respect to $x$
\begin{align*}
\frac{d}{dx} \left(x^{1 / 2}+y^{2 / 3} \right)& = \frac{d}{dx} \left(-4 y\right)\\
\frac{1}{2}x^{-1 / 2}+\frac{2}{3}y^{- 1/ 3} y'(x)&= -4 y'(x)\\
\left (4 + \frac{2}{3}y^{- 1/ 3} \right) y'(x)& = \frac{-1}{2}x^{-1 / 2}\\
\left (\frac{2+12 y^{\frac{1}{3}}}{3 y^{\frac{1}{3}}} \right)y'(x)&= \frac{-1}{2}x^{-1 / 2}\\
\end{align*}
Then
$$y'(x)= -\frac{3 y^{\frac{1}{3}}}{4 \sqrt{x} \cdot\left(1+6 y^{\frac{1}{3}}\right)} $$
