Classify stationary points on two implicit branches

$6x-2y^{\prime}=2yy^{\prime}$, so $y^{\prime}=6x/[2(y+1)]$. Both originals note that there is no need to make $y$ the subject before differentiating. On this real curve $(y+1)^2=1+3x^2>0$, so $y+1\ne0$; no real curve point is lost by this division.

$y^{\prime}=0$ gives $x=0$. Substitute into the original relation: $y^2+2y=0$, hence $y(y+2)=0$ and $y=0$ or $-2$. The candidates are $(0,0)$ and $(0,-2)$, on two different branches.

From $6x-2y^{\prime}=2yy^{\prime}$ obtain $6-2y^{\prime\prime}=2yy^{\prime\prime}+2(y^{\prime})^2$. Thus $y^{\prime\prime}=[6-2(y^{\prime})^2]/[2(y+1)]$. The Teacher note explains why this second derivative test is convenient here, unlike Example 1: the implicit first-derivative equation makes the second derivative relatively easy to obtain.

At both stationary points $y^{\prime}=0$. At $(0,0)$, $y^{\prime\prime}=3>0$, so it is a local minimum. At $(0,-2)$, $y^{\prime\prime}=-3<0$, so it is a local maximum. The source prints the signs and conclusions; these numerical second-derivative values are independently evaluated.

Complete the square: $y^2+2y=3x^2$, $(y+1)^2-1=3x^2$, $(y+1)^2=1+3x^2$, and $y=-1\pm\sqrt{1+3x^2}$. The Teacher blue completion and labelled sketch plot both branches and confirm minimum $(0,0)$ on the upper branch and maximum $(0,-2)$ on the lower. The Student graphing space is blank. Plotting both branches does not turn the whole relation into one single-valued function.