Setup:
Let $z = U(x,y)$
Let $p = \frac{\partial U}{\partial x}$, $q = \frac{\partial U}{\partial y}$
Lagrange's Linear Equation:
$$Pp + Qq = R \tag{1} $$where $P,Q,R$ are polynomials of $x,y,z$
To be proved:
Lagrange's Auxiliary Equations: $$ \frac{dx}{P} = \frac{dy}{Q} =\frac{dz}{R} $$
Method (yields wrong result):
Rewriting equation 1: $$ Pp + Qq - R = 0 \tag{2}$$ Consider vectors $\bar{v}$ and $\bar{u}$
$$\bar{u} = ( p, q, 1 ) \tag{3}$$
$$\bar{v} = ( P, Q, -R ) \tag{4}$$
By equations 2,3,4:
$$(p,q,r).(P,Q,-R)=0 \tag{5}$$
Because $\bar{u}$ = $\nabla u$ : Equation 5: $$\nabla u.\bar{v} = 0 \tag{6}$$
Therefore general solution from 6: Any vector along $\bar{v}$
Therefore by parametric form: $$ \frac{dx}{P} = \frac{dy}{Q} =\frac{dz}{-R} $$
But this is not the same as Lagrange's auxiliary equations. :(