How to verify this identity? From Weinstock, "Calculus of Variations", p.24:

We have the readily verifiable identity
\begin{align}\frac{d}{dx}\left(y'\frac{\partial f}{\partial y'}-f\right) = y'\frac{d}{dx}\left(\frac{\partial f}{\partial y'}\right) - \frac{\partial f}{\partial x} - \frac{\partial f}{\partial y}y'=-y'\left[\frac{\partial f}{\partial y}-\frac{d}{dx}\left(\frac{\partial f}{\partial y'}\right)\right]-\frac{\partial f}{\partial x}\end{align}

Where $f$ is $f(x,y,y')$ and $y$ is a function of $x$.
For some reason I can't seem to verify this identity. Applying the product rule didn't, to the extent of my (admittedly limited) knowledge, seem to help. What am I doing wrong?
 A: You can't prove it- it simply isn't true!
For example, if we take $y'(x)= x^3$ and $f(y')= y'^2$ so that $f(x)= (x^3)^2= x^6$  then $y'\frac{\partial f}{\partial y'}- f= y'(2y')- y'^2= y^2$ so $\frac{d}{dx}\left(y'\frac{\partial f}{\partial y'}- f\right)= \frac{dy^2}{dx}= \frac{dx^6}{dx}= 6x^5$.
But $y'\frac{d}{dx}\left(\frac{\partial f}{\partial y'}\right)-\frac{\partial f}{\partial x}- \frac{\partial f}{\partial y'}y'$ is $(x^3)\frac{d}{dx}(2x^3)- 6x^5- (2x^3)x^3= 6x^5- 6x^5- 2x^6= 2x^6$
A: Thanks to achille hui (and stated in p.5 of the quoted book), I see it now:
If $f$ is $f(x, y, y')$ and $y=y(x)$ then
$$\frac{df}{dx}=\frac{\partial f}{\partial x}+y'\frac{\partial f}{\partial y}+y''\frac{\partial f}{\partial y'}$$
This way,
$$\require{cancel}\begin{align}\frac{d}{dx}\left(y'\frac{\partial f}{\partial y'}-f\right)=y'\frac{d}{dx}\left(\frac{\partial f}{\partial y'}\right)+\cancel{y''\frac{\partial f}{\partial y'}}-\frac{\partial f}{\partial x}-y'\frac{\partial f}{\partial y}-\cancel{y''\frac{\partial f}{\partial y'}}
=y'\frac{d}{dx}\left(\frac{\partial f}{\partial y'}\right)-y'\frac{\partial f}{\partial y}-\frac{\partial f}{\partial x}\end{align}$$
Thereby proving the identity.
