Originally Posted by

**VonNemo19** $\displaystyle x^3+2xy-y^2=-2$

Differentiating with respect to x...

$\displaystyle 3x^2+2x*\frac{dy}{dx}+y*2-2y*\frac{dy}{dx}=0$

$\displaystyle (2x-2y)\frac{dy}{dx}=-3x^2-2y$

$\displaystyle \frac{dy}{dx}=\frac{-3x^2-2y}{2x-2y}$

Now, the points of tangecy are as follows

$\displaystyle (1)^3+2(1)y-y^2=-2$

$\displaystyle y^2-2y-3=0$

$\displaystyle (y-3)(y+1)=0\Rightarrow{y=3,-1}$

Therefore

$\displaystyle \frac{dy}{dx}\text{ at x=1 is }$

$\displaystyle \frac{-3(1)^2-2(3)}{2(1)-2(3)}=\frac{9}{4}$

and

$\displaystyle \frac{-3(1)^2-2(-1)}{2(1)-2(-1)}=\frac{-1}{4}$

So finding the equation...

Is this right so far? is there an easier way to do this?