Chapter 11: Parametric Equations and Polar Coordinates - Section 11.2 - Calculus with Parametric Curves - Exercises 11.2 - Page 657: 19

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Here, we have $\dfrac{dx}{dt}=3t^2+1$ and $\dfrac{dy}{dt}+6t^2=\dfrac{dx}{dt}(2)+2t$ $\implies \dfrac{dy}{dt}=2t+2$ Thus, Slope: $\dfrac{dy}{dx}=\dfrac{2t+2}{3t^2+1}$ or, $\dfrac{2(1)+2}{3(1)^2+1}=\dfrac{4}{4}=1$