Answer
$\dfrac{-3}{16}$
Work Step by Step
Take the derivative of the given equations and isolate the variables.
$3x^2\dfrac{dx}{dt}+4t=0 \implies \dfrac{dx}{dt}=\dfrac{-4t}{3x^2}$
and $6y^2\dfrac{dy}{dt}-6t=0 \implies \dfrac{dy}{dt}=\dfrac{t}{y^2}$
Now, $\dfrac{dy}{dx}=\dfrac{-3x^2}{4y^2}$
Plug the given values of the $x$ and $y$.
we get $x^3+2(2)^2 =9 \implies x=1$
and $2y^3-3(2)^2 =4 \implies y=2$
Thus, $\dfrac{dy}{dx}=\dfrac{-3(1)^2}{4(2)^2}=\dfrac{-3}{16}$
