$\text{(I)}9x^2-6x+1=0$

$\text{(II)}36y^2-12y+1=0$

The correct option is If x > y.

To determine the relationship between $x$ and $y$, we solve both quadratic equations given in the question.

Step 1: Solve Equation (I) for $x$

The given equation is:
$9x^2-6x+1=0$

Notice that this is a perfect square quadratic expression of the form ${\left(a-b\right)}^2=a^2-2ab+b^2$, where $a=3x$ and $b=1$.

Rewriting the equation:
${\left(3x-1\right)}^2=0$

Taking the square root on both sides:
$3x-1=0$
$3x=1$
$x=\frac{1}{3}$

Step 2: Solve Equation (II) for $y$

The given equation is:
$36y^2-12y+1=0$

Similarly, this is also a perfect square expression, where $a=6y$ and $b=1$.

Rewriting the equation:
${\left(6y-1\right)}^2=0$

Taking the square root on both sides:
$6y-1=0$
$6y=1$
$y=\frac{1}{6}$

Step 3: Compare the values of $x$ and $y$

Comparing $x=\frac{1}{3}$ and $y=\frac{1}{6}$:
Since $\frac{1}{3}>\frac{1}{6}$ (or $0.333>0.166$), we have:
$x>y$

Therefore, the correct relationship is If x > y.