If $x^2+xy+3y^2=1\text{ then }(x+6y{)}^3\frac{d^{2}y}{d{x}^2}$ is 

$x^2+xy+3y^2=1$

$\frac{dy}{dx}=\frac{-(2x+y)}{x+6y}$

$\frac{d^{2}y}{d{x}^2}=\frac{\left(x+6y\right)\left(-2-{\displaystyle \frac{dy}{dx}}\right)+\left(2x+y\right)\left(1+6{\displaystyle \frac{dy}{dx}}\right)}{{\left(x+6y\right)}^2}$

${\left(x+6y\right)}^2\frac{d^{2}y}{d{x}^2}=\frac{dy}{dx}\left(12x+6y-x-6y\right)+\left(2x+y-2x-12y\right)$

${\left(x+6y\right)}^2\frac{d^{2}y}{d{x}^2}=\frac{dy}{dx}\left(11x\right)-11y$

${\left(x+6y\right)}^2\frac{d^{2}y}{d{x}^2}=\frac{-(2x+y)}{x+6y}\left(11x\right)-11y$

$$\begin{gathered} \Rightarrow (x+6y{)}^3\frac{d^{2}y}{d{x}^2}=11\left(-2x^2-xy-xy-6y^2\right) \\ \Rightarrow (x+6y{)}^3\frac{d^{2}y}{d{x}^2}=-22\left(x^2+xy+3y^2\right)=-22 \end{gathered}$$