If the tangent at $\left(1,1\right)$  on $y^2=x{\left(2-x\right)}^2$  meets the curve again at p, then p is

$2y\frac{dy}{dx}={\left(2-x\right)}^2-2x\left(2-x\right),\text{so}\frac{dy}{dx}|{}_{\left(1,1\right)}=-\frac{1}{2}.$

Therefore , the equation of tangent at $\left(1,1\right)$ is $y-1=-\frac{1}{2}\left(x-1\right)$

                                                     $\Rightarrow y=\frac{-x+3}{2}$  

The intersection of the tangent and the curve is given by

              $\left(1/4\right){\left(-x+3\right)}^2=x\left(4+x^2-4x\right)$

$\Rightarrow x^2-6x+9=16x+4x^3-16x^2$

$\Rightarrow 4x^3-17x^2+22x-9=0$

$\Rightarrow \left(x-1\right)\left(4x^2-13x+9\right)=0$

$\Rightarrow {\left(x-1\right)}^2\left(4x-9\right)=0$

Since $x=1$  is already the point of tangency , $x=9/4$  and $y^2=\frac{9}{4}{\left(2-\frac{9}{4}\right)}^2=\frac{9}{64}.$

Thus the required point is $\left(9/4,3/8\right).$