The curve _{$y-e^{xy}+x=0$}  has a vertical tangent at the point

_{$\frac{dy}{dx}=\frac{-\left(1-y{e}^{xy}\right)}{1-x.e^{yx}}$}

Since the tangent is a vertical line then _{$1-x.e^{yx}=0\Rightarrow x.e^{yx}=1\Rightarrow x=1,y=0$}