Extraneous and Missing Solution Confusion

After implicit differentiation I came to $$y'x(2xy+1)+y(2xy+1)=0\tag{1}$$ $$(2xy+1)(y'x+y)=0$$

This is correct. Now rather than solving this for $y'$, you could simply substitute $y'=-1$, since this is what you want, and add the equation of the curve to get the following system: $$\left\{\begin{array}{rcl} \left(\color{red}{2xy+1}\right)\left(\color{blue}{y-x}\right) & = & 0 \\ (xy)^2+xy & = & 2 \end{array}\right.$$ You need to consider both equations since you're looking for points:

• on the curve (i.e. satisfying the second equation);
• and with the correct slope (i.e. satisfying the first equation).

Now from the first equation you have either $\color{red}{xy=-\tfrac{1}{2}}$ or $\color{blue}{y=x}$. You say this first case corresponds to an "extraneous solution" but it's not a solution since no points lying on the curve satisfy this equation. Solutions are only those points satisfying both equations.

Substitution of $y=x$ into the second equation leads to $x=\pm 1$ which then leads to the two correct solutions, namely the points $(1,1)$ and $(-1,-1)$.

$$ x^2y^2+xy-2 = 0 $$

Calling $z = xy \Rightarrow z^2+z-2 = (z+2)(z-1) = (xy+2)(xy-1) = 0 \Rightarrow\left\{ \begin{array}{rcl} xy +2 & = & 0\\ xy - 1 & = & 0 \end{array}\right.$

Attached a plot of $(xy+2)(xy-1) = 0$ In red $xy+2=0$ and in blue $xy-1 = 0$

Hence the tangency should be determined over $xy-1=0$ with two solutions for $x = \pm 1$

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