Use implicit differentiation to find an equation of the tangent line to the curve $$x^2+xy+y^2=3, (1,1)$$
I got the derivative as..
$$\frac{2x-2}{x+4}$$
But when I plug in the points I get the equation $y=x/2+2$ which is wrong. Is my derivative wrong? Or am I making a mistake plugging my numbers in. If you could show me where I'm going wrong it'd be much appreciated.
I'm getting a very different derivative. Here's how I did it

$$3=x^2+xy+y^2$$

Now we differentiate

$$0=2x+x \frac{dy}{dx}+y+2y \frac{dy}{dx} $$

Separate terms

$$-(2x+y)=\frac{dy}{dx}(x+2y)$$

so

$$\frac{dy}{dx}=\frac{-(2x+y)}{x+2y}$$

Plugging in $(1,1)$, we get $\frac{dy}{dx}=\frac{-(2+1)}{1+2}=\frac{-3}{3}=-1$, so the tangent line at $(1,1)$, in point-slope form, is $y-1=-(x-1)$, or $y=-x+2$
Ok what about after finding the first derivative I make y the subject in the main function and then substitute the x(1) to find the gradient of the tangent
Please i want a clarification 

solution