I want to show that the Euler equation for the functional $J(y)= \int_a^b f(x,y) \sqrt{1+y'^2}dx$ has the form:
$$f_y-f_xy'-\frac{fy''}{1+y'^2}=0$$
$$L(x,y,y')= f(x,y) \sqrt{1+y'^2} dx$$
Substituting $L_y(x,y,y')=f_y(x,y) \sqrt{1+y'^2}, \ L_{y'}(x,y,y')= f(x,y) \frac{y'}{\sqrt{1+y'^2}}$, I got the following:
$$f_y(1+y'^2)-f_x y'- f y''+ \frac{f (y')^2}{(1+y'^2)}=0$$
Can we get from this relation to the desired one, or have I done something wrong?
The Euler–Lagrange equation is $$ L_y - \frac{dL_{y'} }{dx} = 0. $$ You have $L_y=f_y \sqrt{1+y'^2}$ fine. And $$ L_{y'} = f \frac{y'}{\sqrt{1+y'^2}}. $$ Now we have to take the total derivative of this: $$ \frac{dL_{y'}}{dx} = \frac{y'}{\sqrt{1+y'^2}}\frac{d}{dx} f(x,y) + f \frac{d}{dx} \frac{y'}{\sqrt{1+y'^2}} \\ = \frac{y'}{\sqrt{1+y'^2}} (f_x+y' f_y) + f \frac{y''\sqrt{1+y'^2}+ \frac{y'y''}{\sqrt{1+y'^2}}}{1+y'^2} \\ =\frac{y'}{\sqrt{1+y'^2}} (f_x+y' f_y) + f \frac{y''}{(1+y'^2)^{3/2}}. $$ Therefore the E–L equation is $$ 0 = f_y \sqrt{1+y'^2} - \frac{y'}{\sqrt{1+y'^2}} (f_x+y' f_y) - f \frac{y''}{(1+y'^2)^{3/2}}; $$ multiplying through by $\sqrt{1+y'^2}$ gives $$ 0 = f_y (1+y'^2) - y' f_x - y'^2 f_y - \frac{y''f}{1+y'^2}, $$ which cancels to give the result. Basically, you didn't take the total derivative correctly.