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

The correct option is D $(1,0)$
$\frac{dy}{dx}-e^{xy}[x\frac{dy}{dx}+y]+1=0$
Or
$\frac{dy}{dx}[1-x.e^{xy}]+1-y.e^{xy}=0$
Or
$\frac{dy}{dx}=\frac{-(1-y.e^{xy})}{1-x.e^{xy}}$
Now
$1-x{e}^{xy}=0$ since the slope of the tangent is ${90}^0$.
Hence
$x{e}^{xy}=1$
Or
$x(x+y)=1$
Or
$x^2+xy=1$
Or
$y=\frac{1-x^2}{x}$
Or
$y=\frac{1}{x}-x$ ...(i)
Considering $y=0$,
$x^2=1$
$x=\pm 1$.
Hence we get 2 points $(1,0)$ and $(-1,0)$.
Out of these 2, only $(1,0)$ lies on the curve.
Hence the required point is $(1,0)$.