If $\displaystyle \frac { \cos ^{ 4 }{ \theta } }{ x } +\frac { \sin ^{ 4 }{ \theta } }{ y } =\frac { 1 }{ x+y }$ then $\displaystyle \frac { dy }{ dx } =$

Given $\displaystyle \dfrac { \cos ^{ 4 }{ \theta } }{ x } +\dfrac { \sin ^{ 4 }{ \theta } }{ y } =\dfrac { 1 }{ x+y }$

$\displaystyle \Rightarrow \left( x+y \right) \left( \dfrac { \cos ^{ 4 }{ \theta } }{ x } +\dfrac { \sin ^{ 4 }{ \theta } }{ y } \right) ={ \left( \cos ^{ 2 }{ \theta } +\sin ^{ 2 }{ \theta } \right) }^{ 2 }$ ${ \left( \because \cos ^{ 2 }{ \theta } +\sin ^{ 2 }{ \theta } =1 \right) }$

$\displaystyle \therefore \dfrac { y }{ x } \cos ^{ 4 }{ \theta } +\dfrac { x }{ y } \sin ^{ 4 }{ \theta } -2\sin ^{ 2 }{ \theta } \cos ^{ 2 }{ \theta } =0$

$\displaystyle \Rightarrow { \left( \sqrt { \dfrac { y }{ x } } \cos ^{ 2 }{ \theta } -\sqrt { \dfrac { x }{ y } } \sin ^{ 2 }{ \theta } \right) }^{ 2 }=0$

$\displaystyle \therefore \tan ^{ 2 }{ \theta } =\dfrac { y }{ x } \Rightarrow y=x\tan ^{ 2 }{ \theta }$

$\displaystyle \Rightarrow \dfrac { dy }{ dx } =\tan ^{ 2 }{ \theta }$