Answer: A

Small Hint:

Interpret $\frac{y}{x}$ as the slope of a line through the origin

Big Hint:

At an extreme slope, the line $y=mx$ is tangent to the circle

Solution:

The circle lies entirely where $x\gt0,$ so $\frac{y}{x}$ is the slope $m$ of the line $y=mx$ through a point on it. At an extreme, this line is tangent. The distance from the center $(3,3)$ to $mx-y=0$ must equal $\sqrt6,$ so $$\frac{|3m-3|}{\sqrt{m^2+1}}=\sqrt6.$$ Squaring and simplifying gives $m^2-6m+1=0,$ hence $m=3\pm2\sqrt2.$ The larger value is $3+2\sqrt2.$

Therefore, the correct answer is A.