The curve defined implicitly by $x^3+y^3=6xy$ passes through the point $(3,3)$. (i) Find $\frac{dy}{dx}$ at $(3,3)$ using implicit differentiation. (ii) State whether the tangent at this point is horizontal, vertical, or neither.

Step-by-step solution

1. Differentiate both sides with respect to $x$: $3x^2+3y^2\frac{dy}{dx}=6(y+x\frac{dy}{dx})$.
2. Expand the right side: $6y+6x\frac{dy}{dx}$.
3. Collect the terms containing $\frac{dy}{dx}$ on one side: $3y^2\frac{dy}{dx}-6x\frac{dy}{dx}=6y-3x^2$.
4. Factor $\frac{dy}{dx}$: $\frac{dy}{dx}(3y^2-6x)=6y-3x^2$.
5. Solve for the derivative: $\frac{dy}{dx}=\frac{6y-3x^2}{3y^2-6x}$.
6. Substitute the point $(3,3)$: numerator $=6\cdot 3-3\cdot 3^2=18-27=-9$, denominator $=3\cdot 3^2-6\cdot 3=27-18=9$.
7. Thus $\frac{dy}{dx}=\frac{-9}{9}=-1$ at $(3,3)$.
8. A horizontal tangent would require $\frac{dy}{dx}=0$ and a vertical tangent would require the denominator $3y^2-6x=0$ (making the derivative undefined). Neither condition holds, so the tangent is neither horizontal nor vertical.

Concept tested: Implicit differentiation and tangent classification

Published 7 September 2026 · How this solution was prepared and checked