The locus of centre of a circle which touches externally the circle $x^2+y^2-6x-6y+14=0$  and also touch the $y$  -axis is given by the equation

Let centre of the circle be $\left(h,k\right)$

$\because$  circle touch $y$ -axis

$\therefore$  radius $=\left|h\right|$

Equation of circle is ${\left(x-h\right)}^2+{\left(y-k\right)}^2=h^2$ ………………(i)

Given circle (i) touches externally the circle

$x^2+y^2-6x-6y+14=0$

$\therefore$  Distance between centres $=$ sum of radii

$\sqrt{{\left(h-3\right)}^2+{\left(k-3\right)}^2}=\left|h\right|+2$

$\Rightarrow {\left(h-3\right)}^2+{\left(k-3\right)}^2=h^2+4+4\left|h\right|$

$\Rightarrow k^2-6h-4\left|h\right|-6k+14=0$

Here $h$  always +ve

$\therefore k^2-10h-6k+14=0$

$\therefore$  Locus of centre of circle is

$y^2-10x-6y+14=0$