Answer
$\frac{dy}{dx}=\frac{2y-x^2}{y^2-2x}$
Derivative at $(2,3)$ is $\frac{2}{5}$.
Work Step by Step
$x^3+y^3=6xy-1$
$\frac{d}{dx}[x^3+y^3]=\frac{d}{dx}[6xy-1]$
$3x^2+3y^2\frac{dy}{dx}=6y+6x\frac{dy}{dx}$
$(3y^2-6x)\frac{dy}{dx}=6y-3x^2$
$\frac{dy}{dx}=\frac{6y-3x^2}{3y^2-6x}$
$\frac{dy}{dx}=\frac{2y-x^2}{y^2-2x}$
$(2,3)$
$\frac{2y-x^2}{y^2-2x}=\frac{2(3)-(2)^2}{(3)^2-2(2)}=\frac{2}{5}$
