In the Classical Dynamics of Particles and Systems book, by Stephen T. Thornton and Jerry B. Marion, page 220, the author derived Equation (6.67) from Equation (6.66) which is the following:
Equation (6.67): $$\left(\frac{\partial f}{\partial y} − \ \frac{d}{dx}\frac{\partial f}{\partial y^′}\right)\left(\frac{\partial g}{\partial y}\right)^{−1} = \ \left(\frac{\partial f}{\partial z} − \ \frac{d}{dx}\frac{\partial f}{\partial z^′} \right) \left(\frac{\partial g}{\partial z} \right)^{−1}$$
Because $y$ and $z$ are both functions of $x$, the two sides of Equation (6.67) may be set equal to a function of $x$ which we write as $\lambda(x)$:
Equations (6.68): $$\frac{\partial f}{\partial y} − \frac{d}{dx}\frac{\partial f}{\partial y^′} + \lambda(x)\frac{\partial g}{\partial y} = 0$$ $$\frac{\partial f}{\partial z} − \frac{d}{dx}\frac{\partial f}{\partial z^′} + \lambda(x)\frac{\partial g}{\partial z} = 0$$
where $\lambda(x)$ is the Lagrange undetermined multiplier.
How did the author deduce Equations (6.68) from Equation (6.67)?
Actually, I am a physicist and not a mathematician. I am not an expert in Calculus of Variations. Any help is much appreciated. Thank you so much.