we know that "x" and "y" are encapsulating a function in "t" terms, so then we'll do implicit differentiation on the equation.
[tex]x^2+3y=4\implies \stackrel{chain~rule}{2x\cdot \cfrac{dx}{dt}}+3\cfrac{dy}{dt}=0\implies 2x\cdot \cfrac{dx}{dt}=-3\cfrac{dy}{dt} \\\\\\ \\\stackrel{\textit{"x" with respect to "t"}}{\cfrac{dx}{dt}=-\cfrac{3}{2x}\cdot \cfrac{dy}{dt}}~\hspace{10em}\stackrel{\textit{"y" with respect to "t"}}{-\cfrac{2x}{3}\cdot \cfrac{dx}{dt}=\cfrac{dy}{dt}}[/tex]
