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

Solution:

The equation of the curve is

$\begin{array}{l}y-e^{xy}+x=0\\ \Rightarrow \frac{dy}{dx}-e^{xy}\left(y+x\frac{dy}{dx}\right)+1=0\\ \Rightarrow \frac{dy}{dx}\left(1-x{e}^{xy}\right)=y{e}^{xy}-1\Rightarrow \frac{dx}{dy}=\frac{1-x{e}^{xy}}{y{e}^{xy}-1}\end{array}$

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