The function is not invertible if you think of it as
[tex]f:\mathbb{R}\to\mathbb{R}[/tex]
because it is not injective: [tex]f(x)=f(-x)[/tex]
So, if we restrict the domain to the positive numbers only:
[tex]f:\mathbb{R^+}\to\mathbb{R}[/tex]
Note that the range is actually [tex][9,\infty)[/tex], so we can be more precise and write
[tex]f:\mathbb{R^+}\to[9,\infty)[/tex]
To find the inverse, we solve the expression for x:
[tex]y=x^2+9 \iff y-9=x^2 \iff x=\pm\sqrt{y-9}[/tex]
But since we restricted to positive values of x, the inverse is
[tex]x=\sqrt{y-9}[/tex]