I'm having issues solving $\iiint_v(3x^2+3y^2+3z^2) \, dv$ using Spherical Coordinates
I made the ffg substitutions: $x=r\sin\theta\sin\phi, y=r\sin\theta \cos\phi, z=r\cos\theta$
Thus $3x^2+3y^2+3z^2=3(r^2\sin^2\theta\cos^2\phi+r^2\sin^2\theta\sin^2\phi+r^2\cos^2\theta)=3r^2$
Also $\dfrac{\partial(x,y,z)}{\partial(r,\theta,\phi)}=r^2\sin\theta$
The next step is where I'm not sure if I'm right or wrong:
$$\iiint_v3x^2+3y^2+3z^2\,dv=\iiint3r^2\frac{\partial(x,y,z)}{\partial(r,\theta,\phi)}\,dr\,d\theta \,d\phi=\iiint3r^4\sin\theta \,dr\,d\theta \,d\phi$$
So was I right or wrong?
As pointed out in the comments, you're using $\theta$ as the angle measured for the $z$-axis; $\phi$ as the angle on the $xy$-plane, measured from the positive $x$-axis; and $r$ as the distance from the origin.
Then the conversions from rectangular to spherical coordinates are $$ x=r\cos\phi\sin\theta \\ y=r\sin\phi\sin\theta \\ z=r\cos\phi $$ (I realize this is probably what you meant to write in your question, because this is what you used when you computed $3r^2$)
Your conversion of the function to $3r^2$ looks good, but I think a point needs to be made about the volume element:
You're actually looking for $dV=\bigg|\dfrac{\partial(x,y,z)}{\partial(r,\theta,\phi)}\bigg|\,dr\,d\theta\,d\phi$ (the absolute value is important)
For the Jacobian, you get:
$$\bigg|\dfrac{\partial(x,y,z)}{\partial(r,\theta,\phi)}\bigg|=\big|\,r^2\sin\theta\big|$$
Now, in this case, since $0\leq \theta\leq \pi$, $\sin\theta$ is always nonnegative, so the Jacobian is indeed $r^2\sin\theta$, and
$$dV=r^2\sin\theta \, dr\,d\theta\,d\phi$$
EDIT: To integrate over a sphere (let's say of radius $a$) the limits will be the following: