TITLE: Extraneous and Missing Solution Confusion
QUESTION [9 upvotes]: I came to a question in James Stewart's "Calculus Early Transcendentals" about implicit differentiation saying that:

Find all points on the curve $x^2y^2+xy=2$ where the slope of the tangent line is $-1$

After implicit differentiation I came to $$y'x(2xy+1)+y(2xy+1)=0\tag{1}$$
$$(2xy+1)(y'x+y)=0$$
Then $xy=\frac{-1}{2}$ or $y'=\frac{-y}{x}$
But by pluging $xy=\frac{-1}{2}$ into the original equations we get $$x^2y^2+xy=\frac{1}{4}-\frac{1}{2}\ne2$$
So, we conclude that $xy=\frac{-1}{2}$ is an extraneous solution. Then what caused it to be so ?? Q1

Also, we could start from equation $(1)$ to reach that
$$y'=\frac{-y(2xy+1)}{x(2xy+1)} \tag{2}$$
We need the tangent to be $-1$. So we equate $y'=-1$
$$\frac{y(2xy+1)}{x(2xy+1)}=1$$
We now multiply by $x(2xy+1)$ to get $$y(2xy+1)=x(2xy+1)$$
I am now afraid to divide by $2xy+1$ in order not to get a missing solution.
[Is it wise not to divide ?? Q2 When to remove the common factor of the numerator and denominator and not afraid of missing solutions ?? Q3]
So, I subtracted RHS-LHS to get
$$y(2xy+1)-x(2xy+1)=0$$
Factorizing by grouping
$$(y-x)(2xy+1)=0$$
Now we reached our result $y=x$. Sadly along with the extraneous solution. Is multiplying by the denominator of the rational equation $(2)$ is the cause of the second appearance of the extraneous solution ?? Q4

REPLY [2 votes]: You are over thinking it.
Start with 
$$ (2xy+1)(y'x+y)=0 $$
Set $y'= -1$:
$$ (2xy+1)(-x+y)=0 $$
So $(2xy+1)=0$ and/or $(-x+y)=0$.
$x=y$ is always a solution whether $(2xy+1)=0$ or not.  Since x,y are real, the and condition is never met.
If $(2xy+1)=0$ then $(y'x+y)$ can be any value and doesn't tell you anything about $y'$
Now plug $y=x$ into the original equation to solve for the points.
Hope this helps.
Ced