Finding potential for a vector field

Question

Find the potential function for the vector field $$\frac{16x}z\hat{\mathbf i}+\frac{18y}z\hat{\mathbf j}+\left(1-\frac{8x^2 +9y^2}{z^2}\right)\hat{\mathbf k}$$ which has no constant terms.

So far I have $$f = \frac8z x^2 +\frac9z y^2 +g(z).$$ The $z$ term kinda messes things up and I'm not really able to get any further.

Answer 1

1. Step $1$: $$ f(x,y,z)=\int \frac{16x}{z}\; dx + g(y,z) = \frac{8x^2}{z}+g(y,z) $$
2. Step $2$: $$ \frac{\partial{f(x,y,z)}}{\partial{y}}= \frac{\partial{g(y,z)}}{\partial{y}} = \frac{18y}{z}\quad \Rightarrow \quad g(y,z) = \frac{9y^2}{z}+h(z) $$
3. Step $3$: $$ \frac{\partial{f(x,y,z)}}{\partial{z}}= -\frac{8x^2}{z^2} -\frac{9y^2}{z^2} + h'(z) = (1-\frac{8x^2 +9y^2}{z^2}) \quad \Rightarrow \quad h(z) = z+K\quad (K\in \mathbb{R}) $$ Therefore $$ f(x,y,z)=\frac{8x^2}{z}+\frac{9y^2}{z}+z+K $$ And if there are no constant terms then obviously $K=0$.