The tangent line to the curve has slope equal to [tex]\frac{\mathrm dy}{\mathrm dx}[/tex] at the point (2, -1). We have, by implicit differentiation,
[tex]3y^2-6xy=2x+11\implies6y\dfrac{\mathrm dy}{\mathrm dx}-6y-6x\dfrac{\mathrm dy}{\mathrm dx}=2[/tex]
[tex]\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2+6y}{6y-6x}=\dfrac{1+3y}{3y-3x}[/tex]
At the point (2, -1), the slope is then 2/9, so the tangent line has equation
[tex]y-(-1)=\dfrac29(x-2)\implies y=\dfrac29x-\dfrac{13}9[/tex]
