How do I use spherical coordinates to evaluate $$ \int_0^{3}\int_0^{\sqrt{9-x^2}}\int_0^{\sqrt{9-x^2-y^2}} \frac{\sqrt{x^2+y^2+z^2}}{1+(x^2+y^2+z^2)^2}\, dz\,dy\,dx $$
You don't need to solve it for me, I just want to know how to change the variables to spherical and the setup
The conditions $$\tag{*}0\leq x\leq 3,\ 0\leq y\leq\sqrt{9-x^2-y^2},\ 0\leq z\leq\sqrt{9-x^2-y^2}$$ describe the part of the sphere of radius $3$ that lies on the first octant.
Indeed, the three conditions in $(*)$ clearly imply $$\tag{**}x\geq0,\ y\geq0,\ z\geq0,\ x^2+y^2+z^2\leq 9. $$ Conversely, from $(**)$ we easily get $(*)$.
The sphere of radius $3$, in spherical coordinates, is $0\leq\rho\leq3$. The first octant condition is imposed by $0\leq\theta\leq\pi/2$, $0\leq\phi\leq\pi/2$. So your integral is $$ \int_0^3\int_0^{\pi/2}\int_0^{\pi/2}\frac\rho{1+\rho^4}\,\rho^2\sin\phi\,d\phi\,d\theta\,d\rho =\int_0^3\int_0^{\pi/2}\int_0^{\pi/2}\frac{\rho^3}{1+\rho^4}\,\sin\phi\,d\phi\,d\theta\,d\rho. $$