The curves y = Cx³ form a family, with a different constant C for each curve…

Step 1. The equation of the family. $\dfrac{dy}{dx} = 3Cx^2$ and $C = \dfrac{y}{x^3}$, so $\dfrac{dy}{dx} = \dfrac{3y}{x}$.

Step 2. The swap. $\dfrac{dy}{dx} = -\dfrac{x}{3y}$. Then $3y\,dy = -x\,dx$, so $\dfrac{3y^2}{2} = -\dfrac{x^2}{2} + c$, which is $x^2 + 3y^2 = k$.

Step 3. The point. $1 + 3 \times 4 = 13$, so $k = 13$.

Step 4. The intercept. At $y = 0$, $x^2 = 13$ and $x = 3.6056$.

Step 5. The slope. $-\dfrac{x}{3y} = -\dfrac{1}{6} = -0.1667$ at $(1, 2)$.