Answer
$\frac{dy}{dx}=\frac{-y^2-2xy}{x^2+2xy}$
Derivative at $(-1,1): -1$
Work Step by Step
$(x+y)^3=x^3+y^3$
$x^3+3x^2y+3xy^2+y^3=x^3+y^3$
$3x^2y+3xy^2=0$
$\frac{d}{dx}[3x^2y+3xy^2]=\frac{d}{dx}[0]$
$3(\frac{d}{dx}[x^2y+xy^2])=0$
$\frac{d}{dx}[x^2y+xy^2]=\frac{0}{3}$
$\frac{d}{dx}[x^2y+xy^2]=0$
$2xy+x^2\frac{dy}{dx}+y^2+2xy\frac{dy}{dx}=0$
$2xy+y^2+(x^2+2xy)\frac{dy}{dx}=0$
$(x^2+2xy)\frac{dy}{dx}=-y^2-2xy$
$\frac{dy}{dx}=\frac{-y^2-2xy}{x^2+2xy}$
$(-1,1)$
$\frac{-y^2-2xy}{x^2+2xy}=\frac{-(1)^2-2(-1)(1)}{(-1)^2+2(-1)(1)}=\frac{1}{-1}=-1$
