$\mathrm{If} \frac{{\cos }^4\mathrm{\theta }}{\mathrm{x}}+\frac{{\sin }^4\mathrm{\theta }}{\mathrm{y}}=\frac{1}{\mathrm{x}+\mathrm{y}}\left(\mathrm{\theta } \mathrm{is} \mathrm{cons} \tan \mathrm{t}\right) \mathrm{then} \frac{\mathrm{dy}}{\mathrm{dx}}=$

$$\begin{gathered} \frac{{\cos }^4\mathrm{\theta }}{\mathrm{x}}+\frac{{\sin }^4\mathrm{\theta }}{\mathrm{y}}=\frac{1}{\mathrm{x}+\mathrm{y}}\Rightarrow \frac{{\cos }^4\mathrm{\theta }}{\left({\displaystyle \frac{\mathrm{x}}{\mathrm{x}+\mathrm{y}}}\right)}+\frac{{\sin }^4\mathrm{\theta }}{\left({\displaystyle \frac{\mathrm{y}}{\mathrm{x}+\mathrm{y}}}\right)}={\cos }^2\theta +{\sin }^2\theta \\ \Rightarrow \frac{x}{x+y}={\cos }^2\theta ,\frac{y}{x+y}={\sin }^2\theta \Rightarrow \frac{{\sin }^2\theta }{{\cos }^2\theta }=\frac{y}{x} \\ \Rightarrow y=x{\tan }^2\theta \therefore \frac{dy}{dx}={\tan }^2\theta \left(\sin ce \theta is cons\tan t\right) \end{gathered}$$