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

Differentiating $y-e^{xy}+x=0$ w.r.t. x, we obtain

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

If the curve has a vertical tangent, then $\frac{dy}{dx}=0⇒1-xe^{xy}=0⇒e^{xy}=\frac{1}{x}$

Clearly y = 0, x = 1 satisfies the equations $y -e^{xy} + x = 0$ and $e^{xy}=\frac{1}{x}$