Differentiate both sides with respect to [tex]t[/tex], treating both [tex]x[/tex] and [tex]y[/tex] as functions of [tex]t[/tex].
[tex]x^3 + y^3 = 9 \implies 3x^2 \dfrac{dx}{dt} + 3y^2 \dfrac{dy}{dt} = 0[/tex]
Then when [tex]x=1[/tex], [tex]y=2[/tex] and [tex]\frac{dx}{dt}=-3[/tex], we get
[tex]3\times1^2\times(-3) + 3\times2^2\dfrac{dy}{dt} = 0 \implies \dfrac{dy}{dt} = \dfrac9{12} = \boxed{\frac34}[/tex]
