If \(x^k+y^k=a^k(a, k>0)\) and \(\frac{d y}{d x}+\left(\frac{y}{x}\right)^{\frac{1}{b}}-0\), then \(\mathrm{k}\) has the value

\(x^{\mathrm{k}}+y^{\mathrm{k}}=\mathrm{a}^{\mathrm{k}}\)
Differentiating w.r.t. \(x\), we get
\(\begin{aligned}
& \mathrm{k} x^{k-1}+\mathrm{k} y^{\mathrm{k}-1} \frac{\mathrm{d} y}{\mathrm{~d} x}=0 \\
\therefore \quad & \frac{\mathrm{d} y}{\mathrm{~d} x}=-\frac{\mathrm{k} x^{\mathrm{k}-1}}{\mathrm{k} y^{\mathrm{k}-1}} \\
\therefore \quad & \frac{\mathrm{d} y}{\mathrm{~d} x}=-\left(\frac{x}{y}\right)^{\mathrm{k}-1} \\
& \frac{\mathrm{d} y}{\mathrm{~d} x}+\left(\frac{y}{x}\right)^{1-\mathrm{k}}=0 \\
& \text { But } \frac{\mathrm{d} y}{\mathrm{~d} x}+\left(\frac{y}{x}\right)^{\frac{1}{3}}=0 ...[Given]
\end{aligned}\)
Comparing above equations, we get
\(\begin{aligned}
1-k & =\frac{1}{3} \\
& 1-\frac{1}{3}=k \\
\therefore \quad k & =\frac{2}{3}
\end{aligned}\)