The equation of the common tangent touching the circle  ${\left(x-3\right)}^2+y^2=9$ and the parabola $y^2=4x$ above the x-axis is 

Let the common tangent to the parabola

$y\equiv 4x$  be $y\equiv mx+\frac{1}{m}$ it should also touch the circle ${\left(x-3\right)}^2+y^2=9$ whose centre is at $\left(3,0\right)$ and radius =3, then $\frac{\left|3m+\frac{1}{m}\right|}{\sqrt{1+m^2}}=3\Rightarrow 9m^4+1+6m^2$

$=9m^2\left(1+m^2\right)\Rightarrow 3m^2=1\Rightarrow m=\pm \frac{1}{\sqrt{3}}$

It will make an acute angle with x – axis  $m=\frac{1}{\sqrt{3}}>0$

Equation of common tangent is  $y=\frac{1}{\sqrt{3}}x+\sqrt{3}\Rightarrow \sqrt{3}y=x+3$