Chapter 15 - Section 15.3 - Path Independence, Conservative Fields, and Potential Functions - Exercises - Page 851: 34

$\nabla g=\nabla F=F$ (proved)

Here, $F=\nabla F$ $\int_{0,0,0}^{x,y,z} F \cdot dr=\int_{0,0,0}^{x,y,z} \nabla F dr$ This implies that $dr=f(x,y,z)-f(0,0,0)$ As we know that $\dfrac{\partial g}{\partial x}=\dfrac{\partial f}{\partial x}-0,\dfrac{\partial f}{\partial y}=0$ Now, we have $\nabla g=\nabla F=F$ Hence, proved.