My professor claims that $\nabla^2 \dfrac{\sin(kr)}{kr} = (\partial_r^2+\dfrac{2}{r}\partial_r)\dfrac{\sin(kr)}{kr} = \dfrac{1}{kr}\partial_r^2\sin(kr) = -k\dfrac{\sin(kr)}{r}$ whereas I claim that $\nabla^2 \dfrac{\sin(kr)}{kr}=\dfrac{1}{kr^2}\partial_r(\partial_r r\sin(kr)) = \dfrac{1}{kr^2}\partial_r(\sin(kr)+kr\cos(kr)) = \dfrac{1}{kr^2}[k\cos(kr)+k\cos(kr)-k^2r\sin(kr)] = \dfrac{\cos(kr)}{r^2}-\dfrac{k}{r}\sin(kr)$
Who is right?
You have the $r$s and the derivatives in the wrong order in the Laplacian. There are three ways of writing the radial part in three dimensions: $$ \nabla^2 f(r) = \partial_r^2 f + \frac{2}{r}\partial_r f = f''+\frac{2}{r}f' \\ = \frac{1}{r^2} \partial_r (r^2 \partial_r f) = \frac{1}{r^2} (r^2 f')' \\ = \frac{1}{r} \partial_r^2 (r f) = \frac{1}{r} (rf)'' $$ (these are all easy to check by calculating using the product rule). Your professor has used the last one, while I think you wanted to use the middle one, but got the inside derivative and $r^2$ in the wrong order.