Find horizontal and vertical tangents to a closed curve

The chain rule gives

$\begin{aligned}&2(x+2y)(1+2y^{\prime})\\ &{}+6(x-y)(1-y^{\prime})\\ & =0.\end{aligned}$

Divide by $2$ and expand:

$\begin{aligned}&x+2y+2xy^{\prime}+4yy^{\prime}+3x-3y\\ &{}-3xy^{\prime}+3yy^{\prime}\\ & =0.\end{aligned}$

Hence $4x-y+(7y-x)y^{\prime}=0$, so $(7y-x)y^{\prime}=y-4x$ and $y^{\prime}=(y-4x)/(7y-x)$ wherever the denominator is nonzero.

Set $y-4x=0$, so $x=y/4$. Substitution gives $(9y/4)^2+3(-3y/4)^2=27$, hence $(108/16)y^2=27$ and $y=\pm2$. The tangents are $y=2$ and $y=-2$, at $(1/2,2)$ and $(-1/2,-2)$. Their denominator $7y-x=27y/4$ is nonzero.

Set $7y-x=0$, so $y=x/7$. Then $(9x/7)^2+3(6x/7)^2=27$, giving $(189/49)x^2=27$ and $x^2=7$. Thus the tangents are $x=\sqrt7$ and $x=-\sqrt7$, at $(\sqrt7,1/\sqrt7)$ and $(-\sqrt7,-1/\sqrt7)$. The numerator is $-27x/7\ne0$, so these are regular vertical tangents. The printed notation $dy/dx\to\infty$ denotes vertical slope; a finite real derivative does not exist there.