What are the inverse operations of the "Partial derivative" and the "Total derivative"?

Assume for the moment that $x$, $y$,$z$ are independent variables such that $(x,y,z)$ ranges in an open convex set of ${\mathbb R}^3$.

${\bf 1.\ }$ If you are given a function $(x,y,z)\mapsto f(x,y,z)$ then the indefinite integral $$\int f(x,y,z)\>dx\tag{1}$$ denotes the set of all functions $(x,y,z)\mapsto F(x,y,z)$ satisfying the condition (or PDE) $${\partial F\over\partial x}(x,y,z)=f(x,y,z)\ .\tag{2}$$ If $F_0$ is a solution of $(2)$, found by guessing or by formally integrating $(1)$ with respect to $x$, then the general solution of $(2)$ is given by $$F(x,y,z)=F_0(x,y,z)+G(y,z)\ ,$$ whereby $G$ is an arbitrary (sufficiently smooth) function of its variables $y$ and $z$.

${\bf 2.\ }$ If $(x,y,z)\mapsto F(x,y,z)$ is a scalar function then ${d\over dx}F(x,y,z)$ makes no sense. There is the derivative or differential $dF(x,y,z)$ of $f$, which is for each ${\bf p}=(x,y,z)$ in the domain of $F$ a linear functional on the tangent space $T_{\bf p}$. The geometric representation of this functional is the gradient vector $$\nabla F({\bf p})=\left({\partial F\over\partial x},{\partial F\over\partial y},{\partial F\over\partial z}\right)_{\bf p}\ .$$ Note that $(x,y,z)\mapsto \nabla F(x,y,z)$ constitutes a vector field on the domain in question.

${\bf 3.}\ $Reversing the operation of taking the gradient vector means finding for a given vector field $${\bf f}(x,y,z)=\bigl(u(x,y,z),v(x,y,z),w(x,y,z)\bigr)$$ a scalar function $F$ such that $$\nabla F(x,y,z)={\bf f}(x,y,z)\ .\tag{3}$$ This is not always possible. A necessary condition is that ${\rm curl}({\bf f})\equiv{\bf 0}$. If this condition is fulfilled then you can find an $F$ satisfying $(3)$ either by a recursive scheme ("nested integrals") involving the problem described in ${\bf 1}$, or by computing line integrals as follows: $$F({\bf p})=F({\bf 0})+\int_{\bf 0}^{\bf p}{\bf f}({\bf x})\cdot d{\bf x}\ ,$$ whereby $F({\bf 0})$ is arbitrary.

${\bf 4.\ }$ A "function $f\bigl(x,y(x)\bigr)$", where $x\mapsto y(x)$ is in principle given is a function $$\phi(x):=f\bigl(x,y(x)\bigr)$$ of one variable $x$, and the usual rules of Calculus 101 plus the multivariable chain rule apply. E.g., $$\phi'(x)=f_{.1}\bigl(x,y(x)\bigr)\cdot 1+f_{.2}\bigl(x,y(x)\bigr)\cdot y'(x)\ ,$$ and $$\int f\bigl(x,y(x)\bigr)\>dx=\Phi(x)+C\ ,$$ whereby $\Phi'(x)=\phi(x)$.

Let $f: \mathbb{R^n} \longrightarrow \mathbb{R^p}$ .You can obviously find some integral involving the function $f$ where you want to only consider some of the $(x_1, \cdots, x_n)$ and if that is the case, you assume that the $x_i$ not being used are constants.

For instance let $f(x, y, z) = xyz$. I may want to integrate $f$ with respect to $x$, with $x$ from $0$ to $a$ while keeping $y, z$ constant at, say, $1$. Then I want to compute

$$\int_{0}^{a} xyz\ dx = yz\int_{0}^a x\ dx = \frac{a^2}{2}yz = \frac{a^2}{2}$$

You can also both try to find a function $F$ such that $F\prime = f$ and some function such that $\frac{\partial F}{\partial x_i} = f$ and if you are given the function $f$, that would mean you would be finding a "global" primitive of $f$ or a partial primitive (with respect to $x_i$) of $f$ and thus those operations are well-defined.

For instance finding $F$ such that $\frac{\partial F}{\partial x} = xyz$ would mean finding the primitive with respect to $x$ and in this case is $F = \frac{x^2yz}{2}$

If you want to find $F$ such that $F\prime = f$ you assume that the coordinate functions of $f$ are the partial derivatives of $F$, integrate each one of them, and try to "glue" them together.

Say $f(x, y) = (x^2, y^2)$. Then this means $\frac{\partial F}{\partial x} = x^2$, $\frac{\partial F}{\partial y} = y^2$. Integrating both of them with respect to the right variable you get

$$\frac{\partial F}{\partial x} = x^2 \iff F = \frac{x^3}{3} + \omega_1(y)$$ for some function $\omega_1$ that depends only of $y$.

$$\frac{\partial F}{\partial y} = y^2 \iff F = \frac{y^3}{3} + \omega_2(x)$$ for some function $\omega_2$ that depends only of $x$.

Gluing everything together you get

$$\frac{y^3}{3} + \omega_2(x) = \frac{x^3}{3} + \omega_1(y) \iff \omega_1(y) = \frac{y^3}{3} \wedge \omega_2(x) = \frac{x^3}{3} \rightarrow F(x, y) = \frac{x^3}{3} + \frac{y^3}{3}$$

and getting that

$$\nabla F = f$$ (Please note that I am just saying you can perform such operations, I am not giving you any nomenclature for them)