If you write the equation as $(e^{xy}u_y)_x = 0$, then the boundary conditions tell you that $e^{xy} u_y = e^0u_y(0,y) = -\sin y$, so $$ u_y(x,y) = -\sin y\,e^{-xy}, $$ so the solution is $$ u(x,y) = e^x - \int_0^y\sin t\,e^{-xt}\,dt = e^x + {\frac {{{\rm e}^{-xy}} \left( x\,\sin y +\cos y \right) - 1 }{{x}^{2}+1}}. $$

If you write the equation as $(e^{xy}u_y)_x = 0$, then the boundary conditions tell you that $e^{xy} u_y = e^0u_y(0,y) = -\sin y$, so $$ u_y(x,y) = -\sin y\,e^{-xy}, $$ so the solution is $$ u(x,y) = e^x - \int_0^y\sin t\,e^{-xt}\,dt = e^x + {\frac {{{\rm e}^{-xy}} \left( x\,\sin y +\cos y \right) - 1 }{{x}^{2}+1}}. $$

Remark: If you don't already know about the theory of Laplace invariants of linear second order equations in the plane , you might want to check that out, in case you run into similar problems in the future.