By implicit differentiation, we have

$$4x^3 dx + 4ky^3dy = 0 \implies x^3 + ky^3y' = 0$$

and

$$3x^2dx + k(3y^2y'dy + y^3dy')=0\implies 3x^2 + k(3y^2(y')^2 + y^3 y'') = 0$$

Assuming $y'(2)=y''(2)=k$. Pluging $x=2$, $y=1$ and $y'=y''=k$ in the above two relations yields

$$\left\{\begin{array}{rl} 8 + kn &=0 \\ 12 + k(3n^2 + n)&=0  \end{array}\right. \implies (n, k) = \left(\frac{1}{6},\boxed{-48}\right) $$