Let the equation of the curve by $y=f(x)$ and
curve passes through the point $(1,1)$, So, $f(1)=1$
Also, we know that at any point $(x, y)$ on the curve, the product of the slope of its tangent and $x$ co-ordinate of the point is equal to the $y$ co-ordinate of the point.
$\begin{aligned} \frac{y}{x} & =\frac{d y}{d x} \\ x y^{\prime} & =y\end{aligned}$
$\frac{y^{\prime}}{y}=\frac{1}{x}$
On integrating both sides w.r.t. $x$, we get
$\begin{aligned} \ln |y| & =|\operatorname{In}| x \mid+e^c \\ y & =k x, \text { where } k=e^c\end{aligned}$
Now, we can use the fact that the curve passes through the point $(1,1)$ to find the value of $k$.
$\Rightarrow \mathrm{l}=k(\mathrm{l}) \quad k=1$
Therefore, the equation of the curve is $y=x$ Substituting the value of $x$ and $y$ from the given option, we find that the curve passes through the point $(2,2)$