Computation of determinant for Using Inverse Function Theorem Let $f : \Bbb R^{3} \setminus \{(0, 0, 0)\} → \Bbb R^{3} \setminus \{(0, 0, 0)\}$ be given by
$f(x, y, z) = (x/(x^{2} + y^{2} + z^{2}), y/(x^{2} + y^{2} + z^{2}), z/(x^{2} + y^{2} + z^{2}))$. Show that f is locally invertible at every point in $\Bbb R^{3}\setminus \{(0, 0, 0)\}$. Find an explicit formula for $f^{−1}$.

My attempt:

$$DF(x,y,z)= 
\begin{bmatrix}
-x^{2}+y^{2}+z^{2}/(x^{2}+y^{2}+z^{2})^{2} &amp; -2xy/(x^{2}+y^{2}+z^{2})^{2} &amp; -2xz/(x^{2}+y^{2}+z^{2})^{2}\\
-2xy/(x^{2}+y^{2}+z^{2})^{2}  &amp; x^{2}-y^{2}+z^{2}/(x^{2}+y^{2}+z^{2})^{2} &amp; -2yz/(x^{2}+y^{2}+z^{2})^{2}\\
-2xz/(x^{2}+y^{2}+z^{2})^{2} &amp; -2yz/(x^{2}+y^{2}+z^{2})^{2} &amp; x^{2}+y^{2}-z^{2}/(x^{2}+y^{2}+z^{2})^{2}\\
\end{bmatrix}.
$$   

Calculation of the determinant is computationally difficult here. Is there any other way to solve for this?
Using copper.hat's comment you can easily compute $f^{-1}$ explicitly.

Concerning the Jacobian $J_f$ you can argue as follows: The map $f$ is spherically symmetric, whence $J_f$ depends on $r:=\sqrt{x^2+y^2+z^2}$ only. Since you already have computed the matrix $[df(x,y,z)]$ it is sufficient to look at $$[df(x,0,0)]={\rm diag}\left(-{1\over x^2},{1\over x^2},{1\over x^2}\right)\ .$$
It follows that $J_f(x,0,0)=-{1\over x^6}$, which then implies
$$J_f(x,y,z)=-{1\over r^6}=-{1\over (x^2+y^2+z^2)^3}\ .\tag{1}$$
If you want to avoid computing the matrix $[df(x,y,z)]$ you can consider the spherical shell
$$S:=B(r+h)\setminus B(r)$$ of volume
$${\rm vol}(S)={4\pi\over3}\bigl((r+h)^3-r^3\bigr)\ .$$
Its image $$S':=f(S)=B(1/r)\setminus B\bigl(1/(r+h)\bigr)$$ has volume
$${\rm vol}(S')={4\pi\over3}\left({1\over r^3}-{1\over(r+h)^3}\right)\ .$$
It follows that
$${{\rm vol}(S')\over {\rm vol}(S)}={1\over r^3(r+h)^3}\ \to{1\over r^6}\qquad(h\to 0+)\ .$$
This says that the volume dilatation at radius $r$ is given by ${1\over r^6}$. Since $f$ obviously reverses the orientation we obtain $(1)$.
Here is a complement to Christian's rather nice geometric argument:

Let $n(p) = \|p\|^2$, then $Dn(p) = 2 p^T$.

Let $i(p) = p$, then $Di(p) = I$.

Then $f(p) = (n(p))^{-1} i(p)$, and so, using the product, rule we have
\begin{eqnarray}
Df(p) &amp;=&amp; - (n(p))^{-2} Dn(p) i(p) +  (n(p))^{-1} Di(p) \\
&amp;=&amp; {1 \over \|p\|^2}(I - {2 \over \|p\|^2} p p^T)
\end{eqnarray}
Now, using the fact that $\det(I+AB) = \det (I+BA)$, and $\det (\alpha A) = \alpha^n \det A$, we have
$\det Df(p) = {1 \over (\|p\|^2)^3} (1 - {2 \over \|p\|^2}p^T p) = -{1 \over (\|p\|^2)^3} $.