The given equation of circle is

$x^2+y^2-6x=0\Rightarrow (x-3{)}^2+(y-0{)}^2=(3{)}^2$                       $\dots \left(1\right)$

Here, centre is (3, 0) and radius = 3

Let equation of tangent to the parabola $y^2=4x$ is $y=mx+\frac{1}{m}$                        $\dots \left(2\right)$

$\Rightarrow m^{2}x-my+1=0$ which is also tangent to (i) so, distance from point (3, 0) to the tangent = Radius

$\therefore \left|\frac{3m^2+1}{\sqrt{m^4+m^2}}\right|=3\Rightarrow m=\pm \frac{1}{\sqrt{3}}$

$\therefore$Required equation of tangents are

$x+\sqrt{3}y+3=0$ and $x-\sqrt{3}y+3=0$