# Differentiating this implicit expression?

I am given:

$$\tfrac{1}{4}(x+y)^2 + \tfrac{1}{9}(x - y)^2 = 1$$

Using the chain rule, and factoring out $y'$, I'm left with:

$$y' \left(\tfrac{1}{2}(x+y) - \tfrac{2}{9}(x-y)\right) = 0$$

Now I need to isolate $y'$ but I'm not sure how.

Should I do:

$$y' = \frac{1}{\tfrac{1}{2}(x+y) - \tfrac{2}{9}(x-y)}$$

Thanks

In the implicit differentiation of an equation $f(x,y(x))=C$ you also need to compute the derivative for the x variable, i.e.,
$$0=f_x(x,y(x))+f_y(x,y(x))y'(x)$$
The $f_x$ part is missing in your derivative.
Your first step is wrong. It should be $$\frac{1}{2}(1+y')(x+y)+\frac{2}{9}(1-y')(x-y)=0$$