1. **State the problem:** We are given the differential form $$df = \frac{1}{x^2yz} \, dx + \frac{1}{xy^2z} \, dy + \frac{1}{xyz^2} \, dz$$ and we want to find the function $f(x,y,z)$ such that $df$ is its total differential. 2. **Integrate with respect to $x$:** Treating $y$ and $z$ as constants, integrate the first term: $$f(x,y,z) = \int \frac{1}{x^2 y z} \, dx = \frac{1}{y z} \int x^{-2} \, dx = \frac{1}{y z} \left(-\frac{1}{x}\right) + g(y,z) = -\frac{1}{x y z} + g(y,z)$$ Here, $g(y,z)$ is an arbitrary function of $y$ and $z$. 3. **Differentiate $f$ with respect to $y$ and compare:** $$\frac{\partial f}{\partial y} = \frac{\partial}{\partial y} \left(-\frac{1}{x y z} + g(y,z)\right) = \frac{1}{x y^2 z} + \frac{\partial g}{\partial y}$$ From the given $df$, the coefficient of $dy$ is $$\frac{1}{x y^2 z}$$ Equate: $$\frac{1}{x y^2 z} + \frac{\partial g}{\partial y} = \frac{1}{x y^2 z} \implies \frac{\partial g}{\partial y} = 0$$ So, $g$ does not depend on $y$, i.e., $g(y,z) = h(z)$. 4. **Differentiate $f$ with respect to $z$ and compare:** $$\frac{\partial f}{\partial z} = \frac{\partial}{\partial z} \left(-\frac{1}{x y z} + h(z)\right) = \frac{1}{x y z^2} + h'(z)$$ From the given $df$, the coefficient of $dz$ is $$\frac{1}{x y z^2}$$ Equate: $$\frac{1}{x y z^2} + h'(z) = \frac{1}{x y z^2} \implies h'(z) = 0$$ So, $h(z)$ is a constant. 5. **Conclusion:** The function is $$f(x,y,z) = -\frac{1}{x y z} + C$$ where $C$ is a constant. **Final answer:** $$f(x,y,z) = -\frac{1}{x y z} + C$$