Answer:
We have, \[{{x}^{2/3}}+{{y}^{2/3}}={{a}^{2/3}}\] On differentiating both sides w.r.t. x, we get \[\frac{2}{3}{{x}^{2/3-1}}+\frac{2}{3}{{y}^{2/3-1}}\frac{dy}{dx}=0\] \[\Rightarrow \] \[{{y}^{-1/3}}\frac{dy}{dx}=-{{x}^{^{-1/3}}}\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{-{{x}^{^{-1/3}}}}{{{y}^{^{-1/3}}}}=-\,{{\left( \frac{y}{x} \right)}^{1/3}}\]
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