Answer: \((\frac{1}{2},0)\).

Integrate

\(\frac{dy}{dx}=\frac{12}{(2x+1)^2}.\)

Using \(u=2x+1\), so \(du=2\,dx\),

\(y=\int 12(2x+1)^{-2}\,dx=-\frac{6}{2x+1}+C.\)

The point \((1,1)\) lies on the curve, so

\(1=-\frac{6}{3}+C.\)

Thus

\(C=3.\)

Hence the equation of the curve is

\(y=-\frac{6}{2x+1}+3.\)

At the \(x\)-axis, \(y=0\), so

\(0=-\frac{6}{2x+1}+3.\)

Therefore

\(3=\frac{6}{2x+1}\)

and

\(2x+1=2.\)

So \(x=\frac12\), and the point is

\(\left(\frac12,0\right).\)