Answer
a) $m=1$
b) $y=x$
Work Step by Step
We have: $2x^2+xy=3y^2$
We differentiate both sides with respect to $x$.
$4x+x \dfrac{dy}{dx}+y=6y \dfrac{dy}{dx}\\ \dfrac{dy}{dx}=\dfrac{-4x-y}{x-6y} $
a) The slope at $( -1,-1)$ is: $m=\dfrac{-4(-1)-(-1)}{-1-6(-1)} = 1$
b) The equation of a tangent line at $( -1, -1)$ is:
$y-y_1=m(x-x_1)\\ y+1=1(x+1)\\ y+1=x+1\\ y=x$
