The curve y-e^x y+x=0 has a vertical tangent at the point:

The equation of the curve is

$y-e^{x y}+x=0$

$\Rightarrow \frac{d y}{d x}-e^{x y}\left(y+x \frac{d y}{d x}\right)+1=0$

$\Rightarrow \frac{d y}{d x}\left(1-x e^{x y}\right)=y e^{x y}-1 \Rightarrow \frac{d x}{d y}=\frac{1-x e^{x y}}{y e^{x y}-1}$

Clearly, $\frac{d x}{d y}=0$ at (1, 0). So, required point is (1, 0).