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

Equation of the curve is $y-e^{xy}+x=0$

$⇒\frac{dy}{dx}-e^{xy}(y+x\frac{dy}{dx})+1=0$

$⇒\frac{dy}{dx}(1-xe^{xy})=y.e^{xy}-0$

$⇒\frac{dy}{dx}=\frac{1-xe^{xy}}{y.e^{xy}-1}$

Clearly, $\frac{dy}{dx}=0$ at (1, 0) So, the required point is (1, 0)