Answer
$\dfrac{dx}{dt} = \dfrac {-9}{2}$
Work Step by Step
Given that $\dfrac{dy}{dt} = \dfrac{1}{2}$ and $x^2 y^3=\dfrac{4}{27}$
Then $x^2\dfrac{dy^3}{dt} + y^3\dfrac{d x^2}{dt} = \dfrac{d4}{27dt}$
or, $3x^2y^2\dfrac{dy}{dt} + 2y^3x\dfrac{dx}{dt} =0 $
Now, when $x=2$; then $y^3= \dfrac{4}{27\times 4}$
or, $y=\frac{1}{3}$
$\implies 3\times 4\times\dfrac{1}{18} - 2\times \dfrac{1}{27}(2)\dfrac{dx}{dt}=0$
or, $\dfrac{dx}{dt} = \dfrac {-9}{2}$
