The curve x³ + y³ = 3xy passes through the point (3/2, 3/2)…

Step 1. Differentiate. $3x^2 + 3y^2y' = 3y + 3xy'$.

Step 2. Solve for $y'$. $y'(3y^2 - 3x) = 3y - 3x^2$, so $y' = \dfrac{y - x^2}{y^2 - x}$.

Step 3. Put in the point. $y' = \dfrac{1.5 - 2.25}{2.25 - 1.5} = \dfrac{-0.75}{0.75} = -1$.

Step 4. The tangent. $y - 1.5 = -(x - 1.5)$, which is $y = 3 - x$. It meets the $y$-axis where $x = 0$, at $y = 3$.