I have been asked to show that the Euler-Lagrange of the following fundamental integral $(1)$ can be simplified into $(2)$:
$$ I=\iint_G \sqrt{1+u^2_x(x,y)+u^2_y(x,y)}dxdy \tag 1 $$ Where $u=u(x,y); u_x=\frac{\partial u}{\partial x};u_y=\frac{\partial u}{\partial y}$
$$
u_{xx}(1+u^2_y) + u_{yy}(1+u^2_x)-2u_xu_yu_{xy}=0 \tag 2
$$
I have looked online for help with this question, but the closest I could find is linked below and that does not explain much to me.
Find the surface of least area spanned by a given contour
My calculations thus far are from using the following Equation: $$ \left[\sum \frac{d}{dt_\alpha}\frac{\partial L}{\partial (X_k)_{t_\alpha}}\right]-\frac{\partial L}{\partial u}=0 $$
and this has given me: $$ \frac{d}{dx}\left(\frac{u_x}{\sqrt{1+u^2_x(x,y)+u^2_y(x,y)}}\right)+\frac{d}{dy}\left(\frac{u_y}{\sqrt{1+u^2_x(x,y)+u^2_y(x,y)}}\right)+0=0 $$
Am I on the right track here or am I heading off into a complete different direction?
EDIT: After solving for the complete derivative, I get: $$ \frac{u_{xx}(1+u_y^2)+u_{yy}(1+u_x^2)-2u_xu_yu_{xy}}{\sqrt{1+u_x^2+u_y^2}^3} $$
Is the next step simply stating that because the denominator will never be 0 (actually, the lowest it can be is 1), it is arbitrary and can thus be dropped, leaving us with the numerator? I find this unlikely...
As per your edit you have arrived at the equation
$$\frac{u_{xx}(1+u_y^2)+u_{yy}(1+u_x^2)-2u_xu_yu_{xy}}{\sqrt{1+u_x^2+u_y^2}^3}=0.$$
As you yourself also said, all you have to do now is notice that the denominator is never zero, and so you get an equivalent statement by multiplying by $\sqrt{1+u_x^2+u_y^2}^3$ on both sides. Thus the above is equivalent to
$$u_{xx}(1+u_y^2)+u_{yy}(1+u_x^2)-2u_xu_yu_{xy}=0,$$
which is what you wanted. The key thing here is that you have zero on the right-hand side, which you forgot to write out yourself.