If f is a function of $(x,y,z)$ then the total derivative is
$$d f = \frac { \partial f } { \partial x } d x + \frac { \partial f } { \partial y } d y + \frac { \partial f } { \partial z } d z$$
but what is the derivation of the formula? an user also explains the derivation in a similar question on this platform here
deriving the formula of total derivative of a multi-variable function
But the answer is too complex.Can someone explain this in a simpler way please?thanks .
I'm not sure whether you know what $df$ and the $dx_i$ are. Unless you arrive at the "official vue of things 2019" you cannot understand the formula in your question, let alone its proof. Hence, in the first place you have to have a good understanding of $df$ and the $dx_i$ as mathematical objects!
A differentiable function $$f:\quad {\bf R}^3\to{\mathbb R},\qquad(x_1,x_2,x_3)\mapsto f(x_1,x_2,x_3)$$ has at each point ${\bf p}$ a derivative $df({\bf p})$. This derivative is a linear function $$df({\bf p}):\quad T_{\bf p}\to{\mathbb R},\qquad{\bf X}\to df({\bf p}).{\bf X}\ ,$$ whereby ${\bf X}=(X_1,X_2,X_3)$ is a variable ("small") increment (or tangent) vector attached at ${\bf p}$. This function has the property that it linearly approximates function value increments $f({\bf p}+{\bf X})-f({\bf p})$ as follows: $$f({\bf p}+{\bf X})-f({\bf p})=df({\bf p}).{\bf X}+o(|{\bf X}|)\ \qquad({\bf X}\to{\bf 0})\ .$$ Doing all computations etc. one arrives at the simple formula $$df({\bf p}).{\bf X}={\partial f\over \partial x_1}({\bf p})\>X_1+{\partial f\over \partial x_2}({\bf p})\>X_2+{\partial f\over \partial x_3}({\bf p})\>X_3\ .\tag{1}$$ This formula shows how the linear map $df({\bf p})$ is related to the partial derivatives (or to the Jacobian) of $f$. Now for the coordinate functions $x_i$ $(1\leq i\leq3)$ one has $$x_i({\bf p}+{\bf X})-x_i({\bf p})=X_i$$ without error term. It follows that $$dx_i.{\bf X}=X_i\qquad (1\leq i\leq 3)\>,$$ so that we can rewrite $(1)$ as $$df({\bf p}).{\bf X}={\partial f\over \partial x_1}({\bf p})\ dx_1.{\bf X}+{\partial f\over \partial x_2}({\bf p})\ dx_2.{\bf X}+{\partial f\over \partial x_3}({\bf p})\ dx_3.{\bf X}\ .$$ Since this is true for all tangent vectors ${\bf X}\in T_{\bf p}$ this shows that $$df({\bf p})={\partial f\over \partial x_1}({\bf p})\,dx_1+{\partial f\over \partial x_2}({\bf p})\,dx_2+{\partial f\over \partial x_3}({\bf p})\,dx_3\tag{2}$$ at every point ${\bf p}$ in the domain of $f$. Omitting the ${\bf p}$ the formula $(2)$ then can be abbreviated to $$df={\partial f\over \partial x_1}\ dx_1+{\partial f\over \partial x_2}\ dx_2+{\partial f\over \partial x_3}\ dx_3\ .\tag{3}$$ The formulas $(2)$, resp. $(3)$, say that at each point ${\bf p}$ in the domain of $f$ the derivative $df({\bf p})$ is a certain linear combination of the coordinate differentials $dx_i$, whereby the appearing coefficients are just the partial derivatives of $f$ at ${\bf p}$.
