Given $x^2+xy+y^2=k$

Differentiating the above function,

$2x + y +x\frac{dy}{dx}+2y\frac{dy}{dx}=0$

$\Rightarrow \frac{dy}{dx}=-\frac{\left(y+2x\right)}{x+2y}$

Differentiating the above function,

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

$\Rightarrow \frac{{\mathrm{d}}^{2}y}{d{x}^2}=\frac{-3y + {\displaystyle \frac{\left(-3x\right)\left(y+2x\right)}{\left(x+2y\right)}}}{{\left(x+2y\right)}^2}$

$\Rightarrow \frac{{\mathrm{d}}^{2}y}{d{x}^2}=\frac{-3xy-6y^2-3xy-6x^2}{{\left(x+2y\right)}^3}$

$\Rightarrow \frac{{\mathrm{d}}^{2}y}{d{x}^2}=\frac{-6(x^2+y^2+xy)}{{\left(x+2y\right)}^3}$

$\Rightarrow \frac{{\mathrm{d}}^{2}y}{d{x}^2}=\frac{-6k}{{\left(x+2y\right)}^3}$