x k + y k = a k $\Rightarrow$ kx k - 1 + ky k - 1 ${{{dy} \over {dx}}}$ = 0 $\Rightarrow$ ${{{dy} \over {dx}} + {{\left( {{x \over y}} \right)}^{k - 1}}}$ = 0 ...(1) Given ${{dy} \over {dx}} + {\left( {{y \over x}} \right)^{{1 \over 3}}} = 0$ ...(2) Comparing (1) and (2), we get k - 1 = $- {1 \over 3}$ $\Rightarrow$ k = ${2 \over 3}$