Let $y=y\left(x\right)$ be the solution curve of the differential equation, $(y^2-x)\frac{dy}{dx}=1$, satisfying $y\left(0\right)=1$. This curve intersects the $x$-axis at a point whose abscissa is :

The correct option is C $2-e$
$(y^2-x)\frac{dy}{dx}=1$
$\Rightarrow \frac{dx}{dy}+x=y^2$
$\text{I.F.}=e^{\int 1dy}=e^y$

Therefore, the solution curve is
$x{e}^y=\int y^2{e}^{y}dy$
$\Rightarrow x=y^2-2y+2+c{e}^{-y}$
Given $y\left(0\right)=1$
$\Rightarrow c=-e$
$\therefore$ Solution is $x=y^2-2y+2-e^{-y+1}$
$\therefore$ The value of $x$ where the curve cuts the $x$-axis is $x=2-e$