Finding dy/dx where $x^3+y^3=6xy$

Question

I'm doing a problem where I need to find dy/dx where $x^3+y^3=6xy$.

I got $\frac{x^3+y^3}{6x}= y$

$Fx=\frac{18x^3-6x^3-6y^3}{36x^2}=\frac{2x^3-y^3}{6x^2}$

$Fy=\frac{3y^2x-x^3-y^3}{6x^2}$

and a final answer of $\frac{3y^2x-7y^3}{x^2}$, adding the two partial derivatives together.

The answer is supposed to be $\frac{-x^2+2y}{y^2-2x}$.

Answer 1

Implicit differentiation of $x^3 + y^3 = 6xy$ gives \begin{equation*} \begin{aligned} 3x^2 + 3y^2\frac{dy}{dx} &= 6y + 6x\frac{dy}{dx} \\ \frac{dy}{dx}(3y^2 - 6x) &= 6y - 3x^2 \\ \frac{dy}{dx} &= \frac{2y - x^2}{y^2 - 2x}. \end{aligned} \end{equation*}

Answer 2

$$ \frac{d}{dx}\left[x^3 + y^3 = 6xy\right] \implies 3x^2 + 3y^2 \frac{dy}{dx} = 6y + 6x\frac{dy}{dx} $$

Answer 3

Personally, what I prefer is to write the implicit equation as $F=0$, to compute separately the partial derivatives $F'_x$ and $F'_y$ from which $$\frac{dy}{dx}=-\frac{F'_x}{F'_y}$$ In your case $$F=x^3+y^3-6xy=0$$ $$F'_x=3x^2-6y\qquad \qquad F'_y=3y^2-6x $$ $$\frac{dy}{dx}=-\frac{3x^2-6y}{3y^2-6x}=\frac{2y-x^2}{y^2-2x}$$ If you think about coding, this is exactly how you would it.

Answer 4

$x^3+y^3=6xy$

We can find $\dfrac{dy}{dx}$ for this equation by differentiate both the sides w.r.t. $x$

$\implies \dfrac{d}{dx}(x^3+y^3)=\dfrac{d}{dx}6xy$

$\implies \dfrac{d}{dx}x^3\;+\;\dfrac{d}{dx}y^3=6\dfrac{d}{dx}xy$

$\implies 3x^2\;+\;3y^2\dfrac{dy}{dx}=6\left(x\dfrac{dy}{dx}\;+\;y\right)$

$\implies 3x^2\;+\;3y^2\dfrac{dy}{dx}=6x\dfrac{dy}{dx}\;+\;6y$

$\implies 3y^2\dfrac{dy}{dx}\;-\;6x\dfrac{dy}{dx}=6y\;-\;3x^2$

$\implies (3y^2-6x)\dfrac{dy}{dx}=6y\;-\;3x^2$

$\implies \dfrac{dy}{dx}=\dfrac{6y\;-\;3x^2}{3y^2-6x}$

Thus, $\dfrac{dy}{dx}=\dfrac{2y-x^2}{y^2-2x}$