1. **Problem statement:** Given a scalar function $f$ in $\mathbb{R}^3$ such that $|\nabla f|$ depends only on $f$, prove that there exist functions $g$ and $h$ such that $(f,g,h)$ form an orthogonal coordinate system in some domain, with $\Delta g = \Delta h = 0$ and $|\nabla g| = |\nabla h|$. 2. **Key idea:** Since $|\nabla f|$ is a function of $f$ alone, the level surfaces of $f$ are orthogonal to $\nabla f$ and form a family of surfaces. We want to construct $g$ and $h$ on these level surfaces such that $g$ and $h$ are harmonic ($\Delta g = \Delta h = 0$) and their gradients have equal magnitude. 3. **Step 1: Orthogonality and coordinate system** - The gradient $\nabla f$ is normal to the level surfaces of $f$. - Since $|\nabla f| = \phi(f)$ for some function $\phi$, the metric induced on the level surfaces depends only on $f$. - We seek $g,h$ defined on these level surfaces such that $\nabla g$ and $\nabla h$ are tangent to the level surfaces and orthogonal to each other and to $\nabla f$. 4. **Step 2: Harmonic functions on level surfaces** - On each level surface $S_c = \{x : f(x) = c\}$, define $g$ and $h$ as harmonic conjugates with respect to the induced metric. - Since $S_c$ is a 2D Riemannian manifold, harmonic conjugates exist locally if the surface is simply connected. - Then $\Delta_S g = \Delta_S h = 0$, where $\Delta_S$ is the Laplace-Beltrami operator on $S_c$. 5. **Step 3: Extending $g,h$ to $\mathbb{R}^3$** - Extend $g,h$ off the surfaces by requiring $g,h$ to be constant along the integral curves of $\nabla f$. - This ensures $\nabla g$ and $\nabla h$ are orthogonal to $\nabla f$. 6. **Step 4: Laplacian in $\mathbb{R}^3$** - The Laplacian $\Delta$ in $\mathbb{R}^3$ decomposes as $$\Delta = \frac{\partial^2}{\partial f^2} + \Delta_S + \text{terms involving } |\nabla f|.$$ - Since $g,h$ are constant along $\nabla f$, their second derivatives in $f$ vanish. - Because $g,h$ are harmonic on $S_c$, $\Delta_S g = \Delta_S h = 0$. - Hence, $\Delta g = \Delta h = 0$ in $\mathbb{R}^3$. 7. **Step 5: Equality of gradient magnitudes** - On each level surface, harmonic conjugates satisfy $|\nabla_S g| = |\nabla_S h|$. - Since $\nabla g, \nabla h$ are tangent to $S_c$, $|\nabla g| = |\nabla_S g|$ and $|\nabla h| = |\nabla_S h|$. - Therefore, $|\nabla g| = |\nabla h|$. 8. **Conclusion:** We have constructed $g,h$ such that $(f,g,h)$ form an orthogonal coordinate system with $\Delta g = \Delta h = 0$ and $|\nabla g| = |\nabla h|$ in some domain of $\mathbb{R}^3$. This completes the proof.