Equation of circle touching $x$-axis in origin is ${\left(x-0\right)}^2+{\left(y-k\right)}^2=k^2$
$\Rightarrow x^2+y^2-2ky=0 ...\left(1\right)$, where $k$ is the radius of the circle.

Differentiating the above equation w.r.t. $x$, we get, 

$2x+2y\frac{dy}{dx}-2k\frac{dy}{dx}=0$

$\Rightarrow x+\left(y-k\right)\frac{dy}{dx}=0$

From equation $\left(1\right)$,

$\Rightarrow x+\left(y-\frac{x^2+y^2}{2y}\right)\frac{dy}{dx}=0$

$\Rightarrow \left(x^2-y^2\right)\frac{dy}{dx}=2xy=g\left(x\right)y$

$\Rightarrow g\left(x\right)=2x$