implicit differentiation $\frac1y+\frac1x=1$

Question

I am comparing two approaches to implicit differentiation of the below

$$\frac1y+\frac1x=1$$

The first is:

$$-\frac{dy}{dx}\cdot y^{-2} - x^{-2} = 0$$

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

The second is:

$$1 + \frac yx = y\quad \text{(I multiplied through by} \,y)$$

$$y - \frac yx = 1$$

$$\frac{dy}{dx} - \left[\frac{dy}{dx} \cdot x^{-1} - y \cdot x^{-2}\right] = 0$$

$$\frac{dy}{dx} \cdot (1 - x^{-1}) = -y \cdot x^{-2}$$

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

Can someone please tell me where my error is to arrive at two different answers? Thank you for your time.

Answer 1

Your two answers are the same:

$$y=\frac 1{1-\frac 1x}=\frac x{x-1}\implies y^2=\frac {x^2}{(x-1)^2}\implies -\frac {y^2}{x^2}=-\frac 1 {(x-1)^2}$$

On the other hand $$-\frac y{x(x-1)}=-\frac x{x-1}\times \frac 1{x(x-1)}=-\frac 1{(x-1)^2}$$

Answer 2

$$\frac{dy}{dx} = -\frac y{x^2-x}$$ $$y' = -\frac y{x(x-1)}$$

But you have that $$x+y=yx \implies \frac xy=(x-1)$$ Therefore $$y' = -\frac y{x(x-1)} \implies y' = -\frac 1 {(x-1)^2}=-\frac {y^2}{x^2}$$