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Use implicit differentiation to find dy/dx for x^3+y^3+x+y=1

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the assumption being that "y" is encapsulating a function in "x-terms", whilst "x" is just a simple variable.


x^3+y^3+x+y=1\implies 3x^2+\stackrel{ \textit{chain rule} }{3y^2\cdot \cfrac{dy}{dx}}+1+\cfrac{dy}{dx}=0 \\\\\\ 3y^2\cdot \cfrac{dy}{dx}+\cfrac{dy}{dx}=-3x^2-1\implies \cfrac{dy}{dx}(3y^2+1)=-3x^2-1 \\\\\\ ~\hfill~ {\Large \begin{array}{llll} \cfrac{dy}{dx}=\cfrac{-3x^2-1}{3y^2+1} \end{array}}~\hfill~

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