Q.$x^2+6xy+y^2=10$, show that $\dfrac{d^2y}{dx^2}=\dfrac{80}{(3x+y)^3}$

Given $x^2+6xy+y^2=10$.

Step 1 — differentiate implicitly: $2x+6y+6xy_1+2yy_1=0$, so $y_1=-\dfrac{2x+6y}{6x+2y}=-\dfrac{x+3y}{3x+y}$.

Step 2 — differentiate $2x+6y+6xy_1+2yy_1=0$ again w.r.t. $x$ (product rule on $6xy_1$ and $2yy_1$): $2+6y_1+6y_1+6xy_2+2y_1^2+2yy_2=0$, i.e. $2+12y_1+2y_1^2+(6x+2y)y_2=0$.

Step 3 — solve for $y_2$: $y_2=-\dfrac{2+12y_1+2y_1^2}{6x+2y}=-\dfrac{1+6y_1+y_1^2}{3x+y}$.

Step 4 — let $D=3x+y$, $N=x+3y$, so $y_1=-N/D$. Then $1+6y_1+y_1^2=\dfrac{D^2-6ND+N^2}{D^2}$.

Step 5 — expand $D^2-6ND+N^2=(9x^2+6xy+y^2)-6(3x^2+10xy+3y^2)+(x^2+6xy+9y^2)=-8x^2-48xy-8y^2=-8(x^2+6xy+y^2)$. …

Implicit differentiation twice on $x^2+6xy+y^2=10$, then substitute $y_1=-(x+3y)/(3x+y)$ into the $y_2$ expre …

Forgetting to use the original constraint $x^2+6xy+y^2=10$ to collapse the messy expression $ …