Csdrhrt

i don't know if it's correct but as i remember, the integral of $color{red}{df=dx + dy+dz} $
is: $color{red}{ f=x+y+z}$
and the differential of $color{orange}{f } $ is $color{orange}{d(f) = {partial f over partial x }dx + {partial f over partial y }dy + {partial f over partial z }dz }$ $,,$ so according to the red equation and orange equation, the integral of $d(f)$ should be $color{green}{f={partial f over partial x } + {partial f over partial y } + {partial f over partial z }}$. if we take $f=xyz^2$, with the orange equation df would be $df= yz^2dx + xz^2dy+2xyzdz $ and by the red equation, the f would be $color {blue}{f=xyz^2+xyz^2 + xyz^2 }$ and it is wrong..
i'm super confused..

The red one is correct (except that you forgot the constant of integration), since it can be written as
$$
df = d(x+y+z)
,
$$
which means that $f(x,y,z)=x+y+z+C$.

But the green one is wrong, since the orange one is not (in general) the same thing as
$$
df = dleft( frac{partial f}{partial x} + frac{partial f}{partial y} + frac{partial f}{partial z} right)
.
$$

Integration works a little differently with functions of several variables. You can only integrate with respect to one variable at a time. So, if $f(x, y, z) = x + y + z,$ and the total derivative (differential) here is just the 1-form $$df = dx + dy + dz.$$

But, recovering $f$ is a little different. We cannot just simply integrate, and integrating one at a time doesn't quite do the job, as, for example,

$$ int dx = x + g(y, z) $$

where $g$ is a function of $y$ and $z$ only. This function serves as the "constant" of integration, as taking the derivative with respect to $x$ will make $g$ vanish.

The equation
$$color{orange}{
df = frac{partial f}{partial x},dx + frac{partial f}{partial y},dy + frac{partial f}{partial z},dz
}tag1$$
is true in general; the equation
$$color{red}{df = dx + dy + dz}tag2$$
is a particular special case of Equation $(1)$ in which
$frac{partial f}{partial x} = frac{partial f}{partial y} = frac{partial f}{partial z} = 1.$
And indeed one of the antiderivatives of Equation $(2)$ is $color{red}{f = x + y + z}.$

It is not clear what the green equation,
$color{green}{
f stackrel?= frac{partial f}{partial x} + frac{partial f}{partial y} + frac{partial f}{partial z}},$
was intended to mean.
What it actually does say is easy to disprove;
for example, when
$frac{partial f}{partial x} = frac{partial f}{partial y} = frac{partial f}{partial z} = 1$
the green equation says that
$$
f stackrel?= frac{partial f}{partial x} + frac{partial f}{partial y} + frac{partial f}{partial z}
= 1 + 1 + 1 = 3,
$$
whereas the actual solution for
$frac{partial f}{partial x} = frac{partial f}{partial y} = frac{partial f}{partial z} = 1$
is $f = x + y + z + C,$ not the constant function $f = 3.$
What I suppose you really meant was
$$
f stackrel?= intfrac{partial f}{partial x},dx + intfrac{partial f}{partial y},dy + intfrac{partial f}{partial z},dz,tag3
$$
since that produces your blue equation,
but Equation $(3)$ also is easy to disprove in the general case (as you have just done).
It is easy enough to show that Equation $(3)$ is true is when $f(x,y,z) = f_1(x) + f_2(y) + f_3(z),$
but in the general case we cannot count on being able to express $f$ as a sum of three independent components in that way.

But here is how we might find $f$ in the case where
$df= yz^2 ,dx + xz^2 ,dy + 2xyz ,dz$ as in your example;
that is, where
$frac{partial f}{partial x} = yz^2,$
$frac{partial f}{partial y} = xz^2,$
and $frac{partial f}{partial z} = 2xyz,$
by integrating over each of the variables $x,$ $y,$ and $z.$
We do this by finding the value of $f$ at a particular point, $f(x_1,y_1,z_1).$

First, setting $g(x) = f(x,0,0),$
we find that $g'(x) = left.frac{partial f}{partial x}right|_{y=z=0} = 0,$ and
$$
g(x_1) = f(0,0,0) + int_0^{x_1} g'(x) ,dx
= f(0,0,0) + int_0^{x_1} 0 ,dx = f(0,0,0).
$$
Therefore $f(x_1,0,0) = g(x_1) = f(0,0,0).$

Next, setting $h(y) = f(x_1,y,0),$
we find that $h'(y) = left.frac{partial f}{partial y}right|_{x=x_1,z=0} = 0,$ and
$$
h(y_1) = f(x_1,0,0) + int_0^{y_1} h'(y) ,dy
= f(x_1,0,0) + int_0^{y_1} 0 ,dx = f(x_1,0,0) = f(0,0,0).
$$
Therefore $f(x_1,y_1,0) = h(y_1) = f(0,0,0).$

Finally, setting $k(z) = f(x_1,y_1,z),$
we find that
$k'(z) = left.frac{partial f}{partial z}right|_{x=x_1,y=y_1} = 2x_1y_1z,$ and
begin{align}
k(z_1) &= f(x_1,y_1,0) + int_0^{z_1} k'(z) ,dz\
&= f(x_1,y_1,0) + int_0^{z_1} 2x_1y_1z ,dz\
&= f(x_1,y_1,0) + x_1y_1z_1^2 \
&= f(0,0,0) + x_1y_1z_1^2.
end{align}
Therefore $f(x_1,y_1,z_1) = k(z_1) = f(0,0,0) + x_1y_1z_1^2.$
We can treat $f(0,0,0)$ as the constant of integration, and
obtain the particular antiderivative $f(x,y,z) = xyz^2$ from this result by setting $f(0,0,0) = 0.$

You could also get the same result in the same way by integrating $df$
along a path that starts at $(0,0,0),$ goes straight to $(x,0,0),$
then straight to $(x,y,0),$ and finally straight to $(x,y,z).$