Answer
$\dfrac {1+x\cos xy}{2x-y\cos xy}$
Work Step by Step
$\sin \left( xy\right) =x^{2}-y\Rightarrow \cos \left( xy\right) \left( \dfrac {d}{dx}\left( xy\right) \right) =2x-\dfrac {dy}{dx}\Rightarrow \cos \left( xy\right) \times y+\cos \left( xy\right) \times x\dfrac {dy}{dx}=2x-\dfrac {dy}{dx}\Rightarrow \dfrac {dz}{dx}=\dfrac {1+x\cos xy}{2x-y\cos xy}$
