If the curves

$4 y^{3} \frac{d y}{d x}=1 \quad \& \quad x \frac{d y}{d x}+y=0$

$m_{1}=\frac{1}{4 y^{3}} \frac{d y}{d x}=\frac{-y}{x}=m_{2}$

$m_{1} m_{2}=-1$

$\frac{1}{4 \cdot y^{3}} \times \frac{-y}{x}=-1 \because x=y^{4}$

$\frac{1}{4 \cdot y^{6}}=1$ and $x y=k$

$y^{6}=\frac{1}{4}$

$\Rightarrow k^{6}=y^{30}$

$\Rightarrow k^{6}=\left(\frac{1}{4}\right)^{5}$

$\therefore(4 k)^{6}=4^{6} \times k^{6}=4$