If ${\mathrm{x}}^3+{\mathrm{y}}^3=3{\mathrm{ax}}^2\text{ then }{\mathrm{y}}_2=$

${\mathrm{x}}^3+{\mathrm{y}}^3=3{\mathrm{ax}}^2\Rightarrow {\mathrm{y}}_1=\frac{2\mathrm{ax}-{\mathrm{x}}^2}{{\mathrm{y}}^2}$

$\begin{array}{l}{\mathrm{y}}_2=\frac{{\mathrm{y}}^2(2\mathrm{a}-2\mathrm{x})-\left(2\mathrm{ax}-{\mathrm{x}}^2\right)\cdot 2{\mathrm{yy}}_1}{{\mathrm{y}}^4}=\frac{{\mathrm{y}}^2(2\mathrm{a}-2\mathrm{x})-\left(2\mathrm{ax}-{\mathrm{x}}^2\right)\cdot 2\mathrm{y}\cdot \frac{\left(2\mathrm{ax}-{\mathrm{x}}^2\right)}{{\mathrm{y}}^2}}{{\mathrm{y}}^4}\\ =\frac{2\left[\mathrm{a}\left({\mathrm{x}}^3+{\mathrm{y}}^3\right)-4{\mathrm{a}}^2{\mathrm{x}}^2\right]}{{\mathrm{y}}^5}=\frac{2\mathrm{a}\left[3{\mathrm{ax}}^2-4{\mathrm{ax}}^2\right]}{{\mathrm{y}}^5}=-2\frac{{\mathrm{a}}^2{\mathrm{x}}^2}{{\mathrm{y}}^5}\end{array}$