$f: E^3 \rightarrow E^3$ is an isometry, and $\det f = 1$ and $f'\neq id$

Question

Suppose, that $f: E^3 \rightarrow E^3$ is an isometry, and $\det f = 1$ and $f'\neq id$

Please help me prove, that $f$ is a composition of rotation about an axis and moving along this axis.

I don't know how to start, but tried to write down the matrix and calculate equations for that. I've got equations with 3 degree...

Thanks for all ideas and help!

f' is a linear part of f