If xk + yk = ak(a, k > 0) and \(\frac{\mathrm{d} y}{\mathrm{~d} x}+\left(\frac{y}{x}\right)^{\frac{1}{3}}=0\), then k has the value

1
\(\frac{1}{3}\)
2
\(\frac{2}{3}\)
3
\(\frac{1}{4}\)
4
\(\frac{2}{7}\)

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