$\mathbf{x}\perp f(\mathbf{x})$ and $\|f(\mathbf{x})\|=\|\mathbf{x}\|$ imply jacobian is non-zero

Let $f:\mathbf{R}^3\to \mathbf{R}^3$ be of class $C^1$ and $\mathbf{x}_0\in \mathbf{R}^3$. Suppose there exists an open neighbourhood $U$ of $\mathbf{x}_0$ such that for any $\mathbf{x}\in U$ we have$$\mathbf{x}\cdot f(\mathbf{x})=0$$$$\|f(\mathbf{x})\|_2=\|\mathbf{x}\|_2.$$Prove that the jacobian of $f$ is non-singular in $U$.

I've been given this exercise but don't know how to proceed. The same problem is easy in $\mathbf{R}^2$, where it can be proven that $f(x,y)=(\pm y,\mp x)$ for all $(x,y)\in U$.