Answer
$\dfrac{-3}{16}$
Work Step by Step
Need to take the derivative of the given equations and then isolate the variables.
we have $3x^2\dfrac{dx}{dt}+4t=0$
or, $\dfrac{dx}{dt}=\dfrac{-4t}{3x^2}$
Also, $6y^2\dfrac{dy}{dt}-6t=0$
or, $\dfrac{dy}{dt}=\dfrac{t}{y^2}$
Now, $\dfrac{dy}{dx}=\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}=\dfrac{-3x^2}{4y^2}$
Then, we have $x^3+2(2)^2 =9$; or, $x=1$
and $2y^3-3(2)^2 =4$ or, $y=2$
Plug $x=1,y=2$
Hence, $\dfrac{dy}{dx}=\dfrac{-3(1)^2}{4(2)^2}=\dfrac{-3}{16}$
