Answer
$$y = x$$
Work Step by Step
$$\eqalign{
& 2x{e^{xy}} = {e^{{x^3}}} + y{e^{{x^2}}};\,\,\,\,\,\,\,\,\,\left( {1,1} \right) \cr
& {\text{take the derivative on both sides with respect to }}x \cr
& \frac{d}{{dx}}\left( {2x{e^{xy}}} \right) = \frac{d}{{dx}}\left( {{e^{{x^3}}}} \right) + \frac{d}{{dx}}\left( {y{e^{{x^2}}}} \right) \cr
& {\text{use product rule as follows}} \cr
& 2x\frac{d}{{dx}}\left( {{e^{xy}}} \right) + 2{e^{xy}}\frac{d}{{dx}}\left( x \right) = \frac{d}{{dx}}\left( {{e^{{x^3}}}} \right) + y\frac{d}{{dx}}\left( {{e^{{x^2}}}} \right) + {e^{{x^2}}}\frac{d}{{dx}}\left( y \right) \cr
& {\text{solve the derivatives}} \cr
& 2x{e^{xy}}\left( {x\frac{{dy}}{{dx}} + y} \right) + 2{e^{xy}} = 3{x^2}{e^{{x^3}}} + 2xy{e^{{x^2}}} + {e^{{x^2}}}\frac{{dy}}{{dx}} \cr
& 2{x^2}{e^{xy}}\frac{{dy}}{{dx}} + 2xy{e^{xy}} + 2{e^{xy}} = 3{x^2}{e^{{x^3}}} + 2xy{e^{{x^2}}} + {e^{{x^2}}}\frac{{dy}}{{dx}} \cr
& {\text{combine terms and solve for }}\frac{{dy}}{{dx}} \cr
& 2{x^2}{e^{xy}}\frac{{dy}}{{dx}} - {e^{{x^2}}}\frac{{dy}}{{dx}} = 3{x^2}{e^{{x^3}}} + 2xy{e^{{x^2}}} - 2xy{e^{xy}} - 2{e^{xy}} \cr
& \left( {2{x^2}{e^{xy}} - {e^{{x^2}}}} \right)\frac{{dy}}{{dx}} = 3{x^2}{e^{{x^3}}} + 2xy{e^{{x^2}}} - 2xy{e^{xy}} - 2{e^{xy}} \cr
& \frac{{dy}}{{dx}} = \frac{{3{x^2}{e^{{x^3}}} + 2xy{e^{{x^2}}} - 2xy{e^{xy}} - 2{e^{xy}}}}{{2{x^2}{e^{xy}} - {e^{{x^2}}}}} \cr
& \cr
& {\text{find the slope at the given point }} \cr
& m = {\left. {\frac{{dy}}{{dx}}} \right|_{\left( {1,1} \right)}} = \frac{{3{{\left( 1 \right)}^2}{e^{{{\left( 1 \right)}^3}}} + 2\left( 1 \right)\left( 1 \right){e^{{{\left( 1 \right)}^2}}} - 2\left( 1 \right)\left( 1 \right){e^{\left( 1 \right)\left( 1 \right)}} - 2{e^{\left( 1 \right)\left( 1 \right)}}}}{{2{{\left( 1 \right)}^2}{e^{\left( 1 \right)\left( 1 \right)}} - {e^{{{\left( 1 \right)}^2}}}}} \cr
& m = \frac{{3e + 2e - 2e - 2e}}{{2e - e}} = \frac{e}{e} = 1 \cr
& {\text{find the equation of the tangent line at the point }}\left( {1,1} \right) \cr
& {\text{by using the point - slope form }}y - {y_1} = m\left( {x - {x_1}} \right) \cr
& y - 1 = x - 1 \cr
& y = x \cr} $$
