How do I do Linearization at a point that lies on a curve?

Question

I keep applying the formula to the info given but I keep getting lost/weird answers. Can someone please help me?

I know $L(x)=f(a)+f'(a)(x-a)$

question

Y(x) satisfies $x^2y^2 + xy = 6$. Point (x,y) lies on this curve. Find the linearization of y(x) at x=1.

I tried doing the derivative of the function and plugging in but idk how to get the answer.

I got $y'=2xy^2+2y\frac{dy}{dx}x^2+y+x\frac{dy}{dx}=0$

Please help me. Thanks!

Answer 1

So taking the derivative, $$\begin{split} 0 &= 2xy^2 + 2x^2yy'+xy'+y \\ 0 &= y+ 2xy^2 + \left(2x^2y+x\right)y' \\ y' &= -\frac{y+ 2xy^2}{2x^2y+x} = -\frac{y \left(1+ 2xy\right)} {x\left(2xy+1\right)} = -y/x \\ \end{split} $$ So now you can apply linearization: $$ L(a) = y(a) + (x-a) y'(a) = y(a) - (x-a) y(a)/a $$ and you should be able to take it from here