Absolutely Spherical!

Upon observation, $f(x,y,z) \ge 0$ for all $x,y,z \in \mathbb{R}$ . The minimum value, $m$ , will occur at any of the points $(x,y,z) = (\pm\sqrt{5},0,0); (0,\pm\sqrt{5}, 0); (0,0,\pm\sqrt{5}) \Rightarrow m = \sqrt{5}.$ To find the maximum, $M$ , we can use LaGrange Multipliers according to:

$grad(f) = \lambda \cdot grad(g)$

$\frac{x}{|x|} = \lambda \cdot 2x;$ $\frac{y}{|y|} = \lambda \cdot 2y;$ $\frac{z}{|z|} = \lambda \cdot 2z$

which yields $\frac{1}{|x|} = \frac{1}{|y|} = \frac{1}{|z|} \Rightarrow |x| = |y| = |z|$ ;

and produces $3(|x|)^2 = 3x^2 = 5 \Rightarrow |x| = |y| = |z| = \sqrt{\frac{5}{3}};$

and finally results in $M = 3 \cdot \sqrt{\frac{5}{3}} = \sqrt{15}.$

Calculating $\frac{M}{m}$ produces $\frac{\sqrt{15}}{\sqrt{5}} = \boxed{\sqrt{3}}.$

Just to put things more rigorously, we begin by proving that $\sqrt{5}$ is indeed the minimum. Here I give two proofs.

The first one

Clearly, we can only work in the first quadrant wlog but if you wish, you can work with all 3 dimensions.

$|x|+|y| \geq \sqrt{x^{2}+y^{2}}$

Now, by triangle inequality,

$|x|+|y|+|z| \geq \sqrt{x^{2}+y^{2}}+|z|$

The expression on the right is exactly the sum of the distance from the origin to the point (x,y,z) traverse obliquely in the xy plane and right up parallel to the z axis, so, by triangle inequality again, we have,

$|x|+|y|+|z| \geq \sqrt{x^{2}+y^{2}}+|z| \geq radius =\sqrt{5}$ $\Box$

The second one

Considering the function long enough, we find that the value we want to extremise is the value of c in the equation $f(x,y,z)=x+y+z=c$ and this is just an equation of a plane with normal vector $\left( \begin{matrix} 1 \\ 1 \\ 1 \end{matrix} \right)$

With this fact in mind, the minimum occurs when this plane is closest to the origin whilst still intersecting the sphere and the maximum occurs when this plane is furthest to the origin whilst still intersecting the sphere. $\Box$

Putting all these together,

$\dfrac{M}{m}=\dfrac{\sqrt{15}}{\sqrt{5}}=\sqrt{3}$