$\sin (xy) + \frac{x}{y} = {x^2} - y$

We have, $\sin (xy) + \frac{x}{y} = {x^2} - y$

On differentiating both sides w.r.t. $x$, we get
$\frac{d}{{dx}}(\sin xy) + \frac{d}{{dx}}\left( {\frac{x}{y}} \right) = \frac{d}{{dx}}{x^2} - \frac{d}{{dx}}y$

$\Rightarrow$ $\cos xy \cdot \frac{d}{{dx}}(xy) + \frac{{y\frac{d}{{dx}}x - x \cdot \frac{d}{{dx}}y}}{{{y^2}}} = 2x - \frac{{dy}}{{dx}}$

$\Rightarrow$ $\cos xy \cdot \left[ {x \cdot \frac{d}{{dx}}y + y \cdot \frac{d}{{dx}} \cdot x} \right] + \frac{{y - x\frac{{dy}}{{dx}}}}{{{y^2}}} = 2x - \frac{{dy}}{{dx}}$

$\Rightarrow$ $x\cos xy \cdot \frac{{dy}}{{dx}} + y\cos xy + \frac{y}{{{y^2}}} - \frac{x}{{{y^2}}}\frac{{dy}}{{dx}} = 2x - \frac{{dy}}{{dx}}$

$\Rightarrow$ $x\cos xy \cdot \frac{{dy}}{{dx}} + y\cos xy + \frac{y}{{{y^2}}} - \frac{x}{{{y^2}}}\frac{{dy}}{{dx}} = 2x - \frac{{dy}}{{dx}}$

$\Rightarrow$ $\frac{{dy}}{{dx}}\left[ {x\cos xy - \frac{x}{{{y^2}}} + 1} \right] = 2x - y\cos xy - \frac{y}{{{y^2}}}$

therefore,$\frac{{dy}}{{dx}} = \left[ {\frac{{2xy - {y^2}\cos xy - 1}}{y}} \right]\left[ {\frac{{{y^2}}}{{x{y^2}\cos xy - x + {y^2}}}} \right]$

$= \frac{{\left( {2xy - {y^2}\cos xy - 1} \right)y}}{{\left( {x{y^2}\cos xy - x + {y^2}} \right)}}$

$\Rightarrow \frac{dy}{dx} = = \frac{{\left( {2xy - {y^2}\cos xy - 1} \right)y}}{{\left( {x{y^2}\cos xy - x + {y^2}} \right)}}$