Stuck on implicit differentiation exercise; inquiring about how to approach it

Question

On exercise 8 on section 3.6 of Stewart’s Second Edition Single-Variable Calculus, it is asking me to find the derivative of y with respect to x by implicit differentiation. The problem it gives me is (I apologize if the images are too blurry or unreadable),

Here’s my work; I tried two ways to approach the problem, but I am unsure whether none, both, or either one contains the correct answer. (Again I apologize if the images are too blurry or unreadable; I am typing this on mobile.)

1st approach: Manipulate the implicit equation to one that is easier to differentiate. “P.R.” means the product rule was used at a specific step, which are clearly separated by horizontal lines.

2nd approach: Differentiate both sides of the implicit equation, and then solve for dy/dx.

I’m not sure which method is correct. Any help will be appreciated.

Answer 1

Multiple ways to do it.

$\sqrt{1+x^2y^2}=2xy$

$\frac{2xy^2+2x^2yy'}{2\sqrt{1+x^2y^2}}=2y+2xy'$

$\frac{xy^2+x^2yy'}{\sqrt{1+x^2y^2}}=\frac{xy^2+x^2yy'}{2xy}=2y+2xy'$

$xy^2+x^2yy'=4xy^2+4x^2yy'$

$y'=\frac{3xy^2}{-3x^2y}=\frac{-y}{x}$

Answer 2

Both answers are correct. You can simplify the result of Method 2 by simply using the given equation for the curve.

$$\frac{dy}{dx} = \frac{2y \color{red}{\sqrt{1+x^2y^2}} - xy^2}{x^2y - 2x\color{red}{\sqrt{1+x^2y^2}}} = \frac{4xy^2-xy^2}{x^2y-4x^2y} = -\frac{y}{x}$$