Answer
$y = x$
Work Step by Step
$$\eqalign{
& y{e^{\sin x}} = x\cos y,{\text{ }}\left( {0,0} \right) \cr
& {\text{Taking the derivative of both sides of the equation with }} \cr
& {\text{respect to }}x \cr
& \frac{d}{{dx}}\left[ {y{e^{\sin x}}} \right] = \frac{d}{{dx}}\left[ {x\cos y} \right] \cr
& {\text{Using the product rule}} \cr
& y\frac{d}{{dx}}\left[ {{e^{\sin x}}} \right] + {e^{\sin x}}\frac{d}{{dx}}\left[ y \right] = x\frac{d}{{dx}}\left[ {\cos y} \right] + \cos y\frac{d}{{dx}}\left[ x \right] \cr
& {\text{Computing derivatives}} \cr
& y\left( {{e^{\sin x}}\cos x} \right) + {e^{\sin x}}\frac{{dy}}{{dx}} = x\left( { - \sin y} \right)\frac{{dy}}{{dx}} + \cos y\left( 1 \right) \cr
& {e^{\sin x}}y\cos x + {e^{\sin x}}\frac{{dy}}{{dx}} = - x\sin y\frac{{dy}}{{dx}} + \cos y \cr
& {\text{Collecting the terms that contains }}\frac{{dy}}{{dx}} \cr
& {e^{\sin x}}\frac{{dy}}{{dx}} + x\sin y\frac{{dy}}{{dx}} = \cos y - {e^{\sin x}}y\cos x \cr
& {\text{Factor and solve for }}\frac{{dy}}{{dx}} \cr
& \left( {{e^{\sin x}} + x\sin y} \right)\frac{{dy}}{{dx}} = \cos y - {e^{\sin x}}y\cos x \cr
& \frac{{dy}}{{dx}} = \frac{{\cos y - {e^{\sin x}}y\cos x}}{{{e^{\sin x}} + x\sin y}} \cr
& \cr
& {\text{Find the slope }}m{\text{ at the point }}\left( {0,0} \right) \cr
& m = {\left. {\frac{{dy}}{{dx}}} \right|_{\left( {0,0} \right)}} = \frac{{\cos \left( 0 \right) - {e^{\sin \left( 0 \right)}}\left( 0 \right)\cos \left( 0 \right)}}{{{e^{\sin \left( 0 \right)}} + \left( 0 \right)\sin \left( 0 \right)}} \cr
& m = \frac{1}{1} \cr
& m = 1 \cr
& {\text{Find the equation of the tangent line at the given point}} \cr
& y - {y_1} = m\left( {x - {x_1}} \right) \cr
& \underbrace {\left( {0,0} \right)}_{\left( {{x_1},{y_1}} \right)} \to x = 0{\text{ and }}{y_1} = 0 \cr
& {\text{Therefore}} \cr
& y - 0 = 1\left( {x - 0} \right) \cr
& {\text{Simplify}} \cr
& y = x \cr} $$
